Probability Test - 11

Let us assume \(\mathrm{X}\) represent the number of times selecting defected articles in a random sample space of given 12 articles

Also, the repeated articles in a random sample space are the Bernoulli trials

Clearly, we have \(\mathrm{X}\) has the binomial distribution where \(\mathrm{n}=12\) and \(\mathrm{p}=10 \%=\frac{1}{ 10}\)

And, \(q=1-p=1-\frac{1 }{10}\)

\(=\frac{9 }{10}\)

\(P(X=x)={ }^{n} C_{x} q^{n-x} p^{x}\)

\(={ }^{12} C_{x}\left(\frac{9}{10}\right)^{12^{-x}} ×\left(\frac{1}{10}\right)^{x}\)

Probability of selecting 9 defective articles

\(={ }^{12} C_{9}\left(\frac{9}{10}\right)^{3}\left(\frac{1}{10}\right)^{9}\)

\(=220×\frac{9^{3}}{10^{3}} × \frac{1}{10^{9}} \)

\(=\frac{22 \times 9^{3}}{10^{11}}\)

Hence, the correct option is (C).

Let \(E_{1}\) be the event that item is produced by \(A, E_{2}\) be the event that item is produced by \(B\) and \(X\) be the event that produced product is found to be defective.

Then \(\mathrm{P}\left(\mathrm{E}_{1}\right)=60 \%=\frac{60}{100}\)

\(=\frac{3}{ 5}\)

\(P\left(E_{1}\right)=40 \%=\frac{40}{100}\)

\(=\frac{2}{5}\)

Also \(P\left(\frac{X}{ E_{1}}\right)=P\) (item is defective given that it is produced by machine \(\left.A\right)=2 \%=\frac{2}{ 100}\)

\(=\frac{1 }{50}\)

And \(\mathrm{P}\left(\frac{\mathrm{X}}{ \mathrm{E}_{2}}\right)=\mathrm{P}\) (item is defective given that it is produced by machine \(\left.\mathrm{B}\right)=1 \%\)

\(=\frac{1}{100}\)

Now the probability that item is produced by B, being given that item is defective, is \(P\left(\frac{E_{2}}{A}\right)\).

By using Bayes' theorem, we have

\(\mathrm{P}\left(\frac{\mathrm{E}_{2}}{\mathrm{A}}\right)=\frac{\mathrm{P}\left(\mathrm{E}_{2}\right) \cdot \mathrm{P}\left(\frac{\mathrm{X}}{\mathrm{E}_{2}}\right)}{\mathrm{P}\left(\mathrm{E}_{1}\right) \cdot \mathrm{P}\left(\frac{\mathrm{X}}{ \mathrm{E}_{1}}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \cdot \mathrm{P}\left(\frac{\mathrm{X}}{\mathrm{E}_{2}}\right)}\)

By substituting the values we get

\(=\frac{\frac{2}{5} \times \frac{1}{100}}{\frac{3}{5} \times \frac{2}{100}+\frac{2}{5} \times \frac{1}{100}}\)

\(=\frac{\frac{2}{5} \times \frac{1}{100}}{\frac{1}{500}(6+2)}=\frac{2}{8}\)

\(\Rightarrow \mathrm{P}\left(\frac{\mathrm{E}_{2}}{\mathrm{A}}\right)=\frac{1}{4}\)

Hence, the correct option is (D).

Let \(\mathrm{E} _1\) be the event that the drawn card is a diamond, \(E_ 2\) be the event that the drawn card is not a diamond, and \(\mathrm{A}\) be the event that the card is lost.

As we know, out of 52 cards, 13 cards are diamond and 39 cards are not diamond.

Then \(P\left(E_{1}\right)=\frac{13}{52}\) and \(P\left(E_{2}\right)=\frac{39}{52}\)

Now, when a diamond card is lost then there are 12 diamond cards out of total 51 cards.

Two diamond cards can be drawn out of 12 diamond cards in \({ }^{12} \mathrm{C}_{2}\) ways.

Similarly, two diamond cards can be drawn out of total 51 cards in \({ }^{51} C_{2}\) ways.

Then probability of getting two cards, when one diamond card is lost, is \(P\left(\frac{A }{E_{1}}\right)\).

Also \(P\left(A E_{f}\right)=\frac{{ }^{12} C_{2}}{{ }^{51} C_{2}}\)

\(\text { Also } P\left(\frac{A}{ E_{1}}\right)=\frac{{ }^{12} C_{2}}{{ }^{51} C_{2}} \)

\(=\frac{12 !}{2 ! \times 10 !} \times \frac{2 ! \times 49 !}{51 !} \)

\(=\frac{12 \times 11 \times 10 !}{2 \times 1 \times 10 !} \times \frac{2 \times 1 \times 49 !}{51 \times 50 \times 49 !} \)

\(=\frac{12 \times 11}{51 \times 50}\)

\(=\frac{22}{425}\)

Now, when not a diamond card is lost then there are 13 diamond cards out of total 51 cards.

Two diamond cards can be drawn out of 13 diamond cards in \({ }^{13} \mathrm{C}_{2}\) ways.

Similarly, two diamond cards can be drawn out of total 51 cards in \({ }^{51} \mathrm{C}_{2}\) ways.

Then probability of getting two cards, when card is lost which is not diamond, is \(P\left(\frac{A}{E_{2}}\right)\).

\(\text { Also } P\left(A E_{2}\right)=\frac{{ }^{13} C_{2}}{^{51} C_{2}}\)

\(=\frac{13 !}{2 ! \times 11 !} \times \frac{2 ! \times 49 !}{51 !} \)

\(=\frac{13 \times 12 \times 11 !}{2 \times 1 \times 10 !} \times \frac{2 \times 1 \times 49 !}{51 \times 50 \times 49 !}\)

\(=\frac{13 \times 12}{51 \times 50}\)

\(=\frac{26}{425}\)

Hence, the correct option is (D).

Given,

\(P ( A \cup B )\)= \(\frac{5}{6}\) 

 \(P ( A \cap B )\) = \(\frac{1} {3}\)

\(P ( B )\) = \(\frac{1}2\)

We know that,

\(P(A \cup B)=P(A)+P(B)-P(A \cap B)\)

\(\frac{5}{6}=P(A)+\frac{1}{2}-\frac{1}{3}\)

\(\Rightarrow P(A)=\frac{5}{6}-\frac{1}{2}+\frac{1}{3}\)

\(\Rightarrow P(A)=\frac{5-3+2}{6}\)

\(\Rightarrow P(A)=\frac{4}{6}\)

\(\Rightarrow P(A)=\frac{2}{3}\)

We know that for independent events,

\(P(A) \cdot P(B)\) = \(P(A \cap B)\)

\(P(A \cap B)\) = \((\frac{2}{3}) \times(\frac{1}{2})\)

=\(\frac{1}{3}\)

This is equal to \(P ( A \cap B )\).

Thus events \(A\) and \(B\) are independent events.

Hence, the correct option is (B).

Let us assume \(X\) represent the number of times of getting 5 in 7 throws of the die

Also, the repeated tossing of a die are the Bernoulli trials

Thus, the probability of getting 5 in a single throw, \(p=\frac{1}{6}\)

And, \(q=1-p\)

\(=1-\frac{1}{6}\)

\(=\frac{5}{6}\)

Clearly, we have \(\mathrm{X}\) has the binomial distribution where \(\mathrm{n}=7\) and \(\mathrm{p}=\frac{1}{6}\)

\(P(X=x)={ }^{n} C_{x} q^{n-x} p^{x}\)

\(={ }^{7} C_{x}\left(\frac{5}{6}\right)^{7^{-x}}×\left(\frac{1}{6}\right)^{x}\)

Probability of getting 5 exactly twice in a die \(=P(X=2)\)

\(={ }^{7} C_{2}\left(\frac{5}{6}\right)^{5} ×\left(\frac{1}{6}\right)^{2} \)

\(=21× \left(\frac{5}{6}\right)^{5} × \frac{1}{36} \)

\(=\left(\frac{7}{12}\right)\left(\frac{5}{6}\right)^{5}\)

\(=\left(\frac{7}{12}\right)×\left(\frac{5}{6}\right)^{5}\)

\(=\left(\frac{7}{12}\right)×\left(\frac{3125}{7776}\right)\)

\(=\left(\frac{21875}{93312}\right)\)

\(=0.23\)

Hence, the correct option is (C).

Let there be x number of defective items in a sample of ten items drawn successively.

Now, as we can see that the drawing of the items is done with replacement. Thus, the trials are Bernoulli trials.

Now, probability of getting a defective item, \(p=\frac{5 }{ 100}\)

\(=\frac{1}{20}\)

Thus, \(q=1-\frac{1}{20}\)

\(=\frac{19}{ 20}\)

We can say that \(x\) has a binomial distribution, where \(n=10\) and \(p=\frac{1}{20}\)

Thus, \(P(X=x)={ }^{n} C_{x} q^{n-x} p^{x}\), where \(x=0,1,2 \ldots n\)

\(={ }^{10} \mathrm{C}_{\mathrm{x}}\left(\frac{19}{20}\right)^{10-\mathrm{x}}\left(\frac{1}{20}\right)^{\mathrm{x}}\)

Probability of getting not more than one defective item \(=\mathrm{P}(\mathrm{X} \leq 1)\)

\(=P(X=0)+P(X=1)\)

\(={ }^{10} C_{0}(\frac{19}{20})^{10}(\frac{1}{20})^{0}+{ }^{10} C_{1}(\frac{19}{20})^{9}(\frac{1}{20})^{1}\)

\(=\left(\frac{19}{20}\right)^{10}+10 \times\left(\frac{19}{20}\right)^{9}\left(\frac{1}{20}\right)^{1}\)

\(=\left(\frac{19}{20}\right)^{9}\left[\frac{19}{20}+\frac{10}{20}\right] \)

\(=\left(\frac{19}{20}\right)^{9} \times\left(\frac{29}{20}\right)\)

Hence, the correct option is (C).

Given,

Probability of getting a sir in a throw of die \(=\frac{1}{6}\)

And, probability of not getting a six \(=\frac{5}{ 6}\)

Let us assume, \(p=\frac{1}{6}\) and \(q=\frac{5}{6}\)

Now, we have

Probability that the 2 sixes come in the first five throws of the die

\(={ }^{5} C_{2}\left(\frac{1}{6}\right)^{2}\left(\frac{5}{6}\right)^{3}\)

\(=\frac{10 \times(5)^{2}}{(6)^{5}}\)

Also, Probability that the six come in the sixth throw \(=\frac{10 \times(5)^{2}}{(6)^{4}} \times \frac{1}{6}\)

\(=\frac{10 \times 125}{(6)^{6}} \)

\(=\frac{625}{23328}\)

Hence, the correct option is (A).

Let us assume \(X\) represent the number of times of getting sixes in 6 throws of a die Also, the repeated tossing of die selection are the Bernoulli trials

Thus, probability of getting six in a single throw of die, \(p=\frac{1}{6}\)

Clearly, we have \(X\) has the binomial distribution where \(n=6\) and \(p=\frac{1 }{6}\)

And, \(q=1-p=1-\frac{1 }{6}\)

\(=\frac{5}{6}\)

\(P(X=x)={ }^{n} C_{x} q^{n-x} p^{x} \)

\(={ }^{6} C_{x}\left(\frac{5}{6}\right)^{6-x}\left(\frac{1}{6}\right)^{x}\)

Probability of throwing at most 2 sixes \(=P(X \leq 2)\)

\(=P(X=0)+p(X=1)+P(X=2)\)

\(={ }^{6} C_{0} ×\left(\frac{5}{6}\right)^{6}+{ }^{6} C_{1} ×\left(\frac{5}{6}\right) 65 ×\left(\frac{1}{6}\right)+{ }^{6} C_{2}\left(\frac{5^{4}}{6}\right)×\left(\frac{1}{6}\right)^{2} \)

\(=1 ×\left(\frac{5}{6}\right)^{6}+6 × \frac{1}{6} ×\left(\frac{5}{6}\right)^{5}+15×\frac{1}{36} ×\left(\frac{5}{6}\right)^{4}\)

\(=\left(\frac{5}{6}\right)^{6}+\left(\frac{5}{6}\right)^{5}+\frac{5}{12} ×\left(\frac{5}{6}\right)^{4} \)

\(=\left(\frac{5}{6}\right)^{4}\left[\left(\frac{5}{6}\right)^{2}+\left(\frac{5}{6}\right)+\left(\frac{5}{12}\right)\right]\)

\(=\frac{70}{36} ×\left(\frac{5}{6}\right)^{4} \)

\(=\frac{35}{18}×\left(\frac{5}{6}\right)^{4}\)

\(=\frac{35}{18}×\left(\frac{5×5×5×5}{6×6×6×6}\right)\)

\(=\frac{35}{18}×\left(\frac{625}{1296}\right)\)

\(=\frac{21875}{23328}\)

= 0.937

 = 0.94 (approx.)

Hence, the correct option is (A).

Sample space contains \(2^{4}=16\) elements.

The favorable events are:

\(\{ H , H , H , T \},\{ H , H , T , H \},\{ H , T , H , H \},\{ T , H , H , H \},\{ H , H , H , H \}\)

There are 5 possibilities,

\(\therefore\) The required probability is \(\frac{5}{16}\).

Hence, the correct option is (D).

Given,

Probability of getting a six in a throw of a die \(=\frac{1}{ 6}\)

Also, the probability of not getting a \(6=\frac{5 }{6}\)

Now, there are three cases from which the expected value of the amount which he wins can be calculated:

(i) First case is that, if he gets a six on his first through then the required probability will be \(\frac{1}{6}\)

\(\therefore\) Amount received by him \(=\) Rs. 1

(ii) Secondly, if he gets six on his second throw then the probability \(=(\frac{5}{6} \times \frac{1 }{6})\)

\(=\frac{5}{36}\)

\(\therefore\) Amount received by him \(=-\) Rs. \(1+\) Rs. 1

\(=0\)

(iii) Lastly, if he does not get six in first two throws and gets six in his third throw then the probability \(=\frac{5}{6} \times \frac{5}{6}\) \(\times \frac{1}{6}\)

\(\therefore\) Amount received by him \(=-\) Rs. \(1-\) Rs. \(1+\) Rs. 1

\(=-1\)

So, expected value that he can win \(=\frac{1}{6}-\frac{25}{216}\)

\(=\frac{(36-25)}{216}\)

\(=\frac{11}{216}\)

Hence, the correct option is (C).

The sample space of the experiment is,

\(S=\)\(\{(1, \mathrm{H}),(1, \mathrm{~T}),(2, \mathrm{H}),(2, \mathrm{~T}),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4, \mathrm{H}),(4, \mathrm{~T}),(5, \mathrm{H}),(5, \mathrm{~T},(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}\)

Let \(A\) be the event that the coin shows a tail and \(B\) be the event that at least one die shows 3.

\( A=\{(1, T),(2, T),(4, T) \cdot(5, T)\}\)

\(B=\{(3,1),(3,2) ,(3,3),(3,4) ,(3,5),(3,6) ,(6,3)\}\)

\(\Rightarrow A \cap B-\phi\)

\(\therefore P(A \cap B)-D\)

Then, 

\(P(B)=P(\{3,1\})+P(\{3,2\})+P(\{3,3\})+P(\{3,4\})+P(13,5\})+P(\{3,6\})+P(\{6,3\})\)

\(=\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}\)

\(=\frac{7}{36}\)

Probability of the event that the coin shows a tail, given that at least one die shows 3 is given by,

\(P(\frac{A}{B})\) = \(\frac{P(A \cap B)}{P(B)}\)

\(=\frac{0}{(\frac{7}{36})}\)

\(=0\)

Hence, the correct option is (A).

Let us assume that, man tosses the coin \(n\) times. Thus, \(n\) tosses are the Bernoulli trials

Probability of getting head at the toss of the coin \(=\frac{1}{2}\)

Let us assume, \(p=\frac{1}{2}\) and \(q=\frac{1}{2}\)

\(P(X=x)={ }^{n} C_{x} p^{n-x} q^{x} \)

\(={ }^{n} C_{x}\left(\frac{1}{2}\right)^{n-x}\left(\frac{1}{2}\right)^{x}\)

\(={ }^{n} C_{x}\left(\frac{1}{2}\right)^{n}\)

It is given in the question that,

Probability of getting at least one head \(>\frac{90}{100}\)

Therefore,

\(P(x \geq 1)>0.9\)

\(1-P(x=0)>0.9 \)

\(1-{ }^{n} C_{0} \cdot \frac{1}{2^{n}}>0.9 \)

\(\frac{1}{2^{n}}<0.1\)

\(2^{n}>\frac{1}{0.1} \)

\(2^{n}>10\)

So, the minimum value of \(n\) satisfying the given inequality \(=4\)

\(\therefore\) The man have to toss the coin 4 or more times.

Hence, the correct option is (A).

Given,

A pack of 52 cards.

As we know there are 26 cards in total which are black. Let A and B denote respectively the events that the first and second drawn cards are black.

Now, P (A) = P (black card in first draw) = \(\frac{26}{52}\) = \(\frac{1}{2}\)

Because the second card is drawn without replacement so, now the total number of black cards will be 25 and the total cards will be 51 that is the conditional probability of B given that A has already occurred.

Now, \(P (\frac{B}{A})\) = P (black card in second draw) = \(\frac{25}{51}\)

Thus the probability that both the cards are black.

= P (A ∩ B) = \(\frac{1}{2}\) × \(\frac{25}{51}\) = \(\frac{25}{102}\)

Therefore, the probability that both the cards are black is \(\frac{25}{102}\).

Hence, the correct option is (A).

Given,

Probability of defective eggs \(=10 \%\)

\(\mathrm{p}=\frac{10}{100}\)

\(=\frac{1}{10}\)

(Probability of good eggs) \(=\mathrm{q}\)

\(\mathrm{q}=1-\mathrm{p}\)

\(=1-\frac{10}{100}\)

\(=\frac{9}{10}\)

(Probability of at least one egg defective out of 10 )

\(\mathrm{p}(1)+\mathrm{p}(2)+\mathrm{p}(3)+\ldots \)

\(=\mathrm{p}(0)+\mathrm{p}(1)+\mathrm{p}(2)+\ldots+\mathrm{p}(10)-\mathrm{p}(0)\)

\(=[\mathrm{p}(0)+\mathrm{p}(1)+\mathrm{p}(2)+\ldots+\mathrm{p}(10)]-\mathrm{p}(0)\)

\(=1-\mathrm{p}(0)\)

\(=1-\left(\frac{9}{10}\right)^{10}\)

Hence, the correct option is (A).

Let the events be defined as:

A: Obtaining a sum of 8

B: Getting an even number on both dice

Now cases favourable to A are \((3,5)(5,3)(2,6)(6,2)(4,4)\)

So, \(P(A)=\frac{5}{36}\)

Cases favourable to B: \((2,2),(2,4),(2,6),(4,2),(4,4),(4,6),(6,2),(6,4),(6,6)\).

P(B)=\(\frac{9}{36}\)

Now, \((2,6)(6,2)\) and \((4,4)\) are common to both events \(A\) and \(B\)

So, \(P(A \cap B)=\frac{3}{36}\)

\(\Rightarrow P ( A \cup B )=\frac{5}{36}+\frac{9}{36}-\frac{3}{36}\)

\(\Rightarrow P ( A \cup B )\)\(=\frac{11}{36}\)

Hence, the correct option is (C).

Let, \(E_{1}\) be the event of choosing the bag 1 \(E_{2}\) be the event of choosing the bag say bag 2 and \(A\) be the event of drawing a red ball.

Also \(P(\frac{A}{E_{1}})=P\) (drawing a red ball from bag 1) = \(\frac{4}{8}\) 

= \(\frac{1}{2}\)

Also \(P(\frac{A}{E_{2}})=P\) (drawing a red ball from bag 2 ) = \(\frac{2}{8}\) 

= \(\frac{1}{4}\)

Now the probability of drawing a ball from bag 1, being given that it is red = \(P(\frac{E_{1}}{A})\).

By using Bayes' theorem, we have,

\(=\frac{P\left(E_{1}\right) \cdot P\left(\frac{A}{E_{1}}\right)}{P\left(E_{1}\right) \cdot P\left(\frac{A}{ E_{1}}\right)+P\left(E_{2}\right) \cdot P\left(\frac{A}{E_{2}}\right)}\)

\(\Rightarrow P ( \frac{E _{1}}{A})\)\(=\frac{(\frac{1}{2} \times \frac{1}{2})}{(\frac{1}{2}\times\frac{1}{2})+(\frac{1}{2}\times\frac{1}{4})}\)

\(\Rightarrow P ( \frac{E _{1}}{A})\)\(=\frac{(\frac{1}{4})}{(\frac{1}{4}+\frac{1}{8})}\)

\(\Rightarrow P ( \frac{E _{1}}{A})\)\(=\frac{(\frac{1}{4})}{(\frac{3}{8})}\)

\(\Rightarrow P ( \frac{E _{1}}{A})\)\(=\frac{2}{3}\)

Hence, the correct option is (B).

Given,

A die is tossed thrice

Then the sample space S = {1, 2, 3, 4, 5, 6}

Let P (A) = probability of getting an odd number in the first throw.

P (A) = \(\frac{3}{6}\) = \(\frac{1}{2}\)

Let P (B) = probability of getting an even number.

⇒ P (B) =  \(\frac{3}{6}\) = \(\frac{1}{2}\)

Now, probability of getting an even number in three times = \(\frac{1}{2} × \frac{1}{2} × \frac{1}{2} = \frac{1}{8}\)

So, probability of getting an odd number at least once

= \(1\) – probability of getting an odd number in no throw

= \(1\) – probability of getting an even number in three times

= \(1\) – \(\frac{1}{8}\)

= \(\frac{7}{8}\)

∴ Probability of getting an odd number at least once is \(\frac{7}{8}\).

Hence, the correct option is (C).

We know that in a leap year there are a total of 366 days, 52 weeks, and 2 days

Now, in 52 weeks there are a total of 52 Tuesdays.

Therefore,

The probability that the leap year will contain 53 Tuesdays is equal to the probability of the remaining 2 days will be Tuesdays. Thus, the remaining two days can be:

(Monday and Tuesday), (Tuesday and Wednesday), (Wednesday and Thursday), (Thursday and Friday), (Friday and Saturday), (Saturday and Sunday), and (Sunday and Monday)

Therefore,

Total Number of cases = 7

Cases in which Tuesday can come = 2

So, probability (leap year having 53 Tuesdays) =\(\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\)

= \(\frac{2}{7}\)

Hence, the correct option is (A).

Given,

The probability of failure \(= x\)

And, probablity of suocess \(=2 x\)

Therefore,

\( x+2 x=1 \)

\(3 x-1 \)

\(x=\frac{1}{3} \)

\(2 x=\frac{2}{3}\)

Assume \(p=\frac{1}{3}\) and \(q=\frac{2}{3}\)

Aleo, \(\mathrm{X}\) be the random variable that represents the number of trials Hence, by binomial distribution we have:

\(P(X=x)={ }^{n} C_{x} p^{n-x} q^{x}\)

\(\therefore\) Probability of having at least 4 successes \(=P(X \geq 4)\)

\(=P(X=4)+P(X=5)+P(X=6) \)

\(=C_{4}\left(\frac{2}{3}\right)^{4}\left(\frac{1}{3}\right)^{2}+^{6} C_{5}\left(\frac{2}{3}\right)^{5}\left(\frac{1}{3}\right)+{ }^{6} C_{6}\left(\frac{2}{3}\right)^{6}\)

\(=\frac{15(2)^{4}}{3^{6}}+\frac{6(2)^{5}}{3^{6}}+\frac{(2)^{6}}{3^{6}} \)

\(=\frac{31 \times(2)^{4}}{(3)^{6}}\)

\(=\frac{31}{9}\left(\frac{2}{3}\right)^{4}\)

Hence, the correct option is (B).

Let us firstly assume, \(\mathrm{A}_{1}\) denote the events that a red ball is transferred from bag \(\mathrm{I}\) to \(\mathrm{II}\)

And, \(\mathrm{A}_{2}\) denote the event that a black ball is transferred from bag \(\mathrm{I}\) to \(\mathrm{II}\)

\(\mathrm{P}\left(\mathrm{A}_{1}\right)=\frac{3}{ 7}\)

And, \(P\left(A_{2}\right)=\frac{4 }{7}\)

Let \(\mathrm{X}\) be the event that the drawn ball is red

when red ball is transferred from bag \(\mathrm{I}\) to \(\mathrm{II}\),

\(P\left(\frac{X }{A_{1}}\right)=\frac{5}{10} \)

\(=\frac{1}{2}\)

And, when black ball is transferred from bag \(\mathrm{I}\) to \(\mathrm{II}\)

\(P\left(\frac{X}{A_{2}}\right)=\frac{4}{10} \)

\(=\frac{2}{5} \)

\(P\left(\frac{A_{2} }{X}\right)=\frac{P\left(A_{2}\right) P\left(\frac{X }{A_{2}}\right)}{P\left(A_{1}\right) P\left(\frac{X}{A_{1}}\right)+P\left(A_{2}\right) P\left(\frac{X }{A_{2}}\right)} \)

\(=\frac{\frac{4}{7} \times \frac{2}{5}}{\frac{3}{7} \times \frac{1}{2}+\frac{4}{7} \times \frac{2}{5}} \)

\(=\frac{16}{31}\)

Hence, the correct option is (D).

Let us now assume, \(\mathrm{X}\) represents the number of correct answers by guessing in the multiple-choice set

Now, probability of getting a correct answer, \(p=\frac{1 }{3}\)

Thus, \(q=1-p=1-\frac{1}{3}\)

\(=\frac{2}{3}\)

Clearly, we have \(\mathrm{X}\) is a binomial distribution where \(\mathrm{n}=5\) and \(\mathrm{P}=\frac{1}{3}\)

\(P(X=x)={ }^{n} C_{x} q^{n-x} p^{x} \)

\(={ }^{5} C_{x}\left(\frac{2}{3}\right)^{5^{-x}} \cdot\left(\frac{1}{3}\right)^{x}\)

Hence, probability of guessing more than 4 correct answer \(=P(X \geq 4)\)

\(=P(X=4)+P(X=5)\)

\(={ }^{5} C_{4}\left(\frac{2}{3}\right) \cdot\left(\frac{1}{3}\right)^{4}+{ }^{5} C_{5}\left(\frac{1}{3}\right)^{5}\)

\(=5 \cdot (\frac{2}{3}) .(\frac{1}{81})+1 \cdot (\frac{1}{243})\)

\(=\frac{11}{243}\)

Hence, the correct option is (C).

Let \(E_{1}\) be the event of time consumed by machine \(A, E_{2}\) be the event of time consumed by machine \(B\) and \(E_{2}\) be the event of time consumed by machine \(C\). Let \(X\) be the event of producing defective items.

Then \(P\left(E_{1}\right)=50 \%=\frac{50}{100}\)

\(=\frac{1}{2}\)

\(P\left(E_{2}\right)=30 \%=\frac{30}{100}\)

\(=\frac{3}{10}\)

\(P\left(E_{3}\right)=20 \%_{b}=\frac{20}{100}\)

\(=\frac{1}{5}\)

As we a headed coin has head on both sides so it will shows head.

Also \(P\left(\frac{X }{E_{1}}\right)=P(\) defective item produced by \(A)=1 \%\)

\(=\frac{1}{100}\)

And \(P\left(\frac{X}{ E_{2}}\right)=P\) (defective item produced by \(\left.B\right)=5 \%\)

\(=\frac{5}{100}\)

And \(P\left(\frac{X}{ E_{3}}\right)=P\) (defective item produced by \(\left.C\right)=7 \%\)

\(=\frac{7}{100}\)

Now the probability that item produced by machine \(A\), being given that defective item is produced, is \(P\left(\frac{E_{1}}{ A}\right)\).

By using Bayes' theorem, we have

\(\mathrm{P}\left(\frac{\mathrm{E}_{1}}{ \mathrm{X}}\right)=\frac{\mathrm{P}\left(\mathrm{E}_{1}\right) \cdot \mathrm{P}\left(\frac{\mathrm{X}}{\mathrm{E}_{1}}\right)}{\mathrm{P}\left(\mathrm{E}_{1}\right) \cdot \mathrm{P}\left(\frac{\mathrm{X}}{\mathrm{E}_{1}}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \cdot \mathrm{P}\left(\frac{\mathrm{X}}{\mathrm{E}_{2}}\right)+\mathrm{P}\left(\mathrm{E}_{3}\right) \cdot \mathrm{P}\left(\frac{\mathrm{X}}{\mathrm{E}_{3}}\right)}\)

Now by substituting the values we get

\(=\frac{(\frac{1}{2}) \cdot (\frac{1}{100})}{(\frac{1}{2}) \cdot (\frac{1}{100})+(\frac{3}{10}) \cdot (\frac{5}{100})+(\frac{1}{5}) \cdot (\frac{7}{100})}\)

\(=\frac{(\frac{1}{2} )\cdot (\frac{1}{100})}{(\frac{1}{100})\left((\frac{1}{2})+(\frac{3}{2})+(\frac{7}{5})\right)}\)

\(=\frac{(\frac{1}{2})}{(\frac{17}{5})}=(\frac{5}{34} )\)

\(\Rightarrow \mathrm{P}\left(\frac{\mathrm{E}_{1}}{\mathrm{X}}\right)=\frac{5}{34}\)

Hence, the correct option is (A).

Given,

\(P(A)=2 P(B)=6 P(C)\)

We know that when events are mutually exclusive and exhaustive,

\(P(A)+P(B)+P(C)=1\)

Let, \(P(A)\) = \(k\)

Then, \(P(B)= \frac{P(A)}{2}=\frac{k}{2}\)

\(P(B)=\frac{k}{2}\)

\(\Rightarrow P(C)= \frac{k}{6}\)

So, according to the concept

\(k+\frac{k}{2}+\frac{k}{6}=1\)

\(\Rightarrow \frac{10 k}{6}=1\)

\(\Rightarrow k=\frac{3}{5}\)

Therefore,

\(P(B)\) = \(\frac{k}{2}=\frac{3}{5 \times 2}\)

\(=0.3\)

Hence, the correct option is (B).

Let \(E_{1}\) be the event that first group wins the competition, \(E_{2}\) be the event that that second group wins the competition and \(\mathrm{A}\) be the event of introducing a new product.

Then \(P\left(E_{1}\right)=0.6\) and \(P\left(E_{2}\right)=0.4\)

Also \(P\left(\frac{A}{ E_{1}}\right)=P\) (introducing a new product given that first group wins) \(=0.7\)

And \(P\left(\frac{A}{E_{2}}\right)=P\) (introducing a new product given that second group wins) \(=0.3\)

Now the probability of that new product introduced was by the second group, being given that a new product was introduced, is \(P\left(\frac{E_{2}}{A}\right)\).

By using Bayes' theorem, we have

\(\mathrm{P}\left(\frac{E_{2}}{ \mathrm{A}}\right)=\frac{\mathrm{P}\left(\mathrm{E}_{2}\right) \cdot \mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{E}_{2}}\right)}{\mathrm{P}\left(\mathrm{E}_{1}\right) \cdot \mathrm{P}\left(\frac{\mathrm{A}}{ \mathrm{E}_{1}}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \cdot \mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{E}_{2}}\right)}\)

Now by substituting the values we get

\(=\frac{0.4 \times 0.3}{0.6 \times 0.7+0.4 \times 0.3}\)

\(=\frac{0.12}{0.42+0.12}\)

\(=\frac{0.12}{0.54}\)

\(=\frac{12}{54}\)

\(=\frac{2}{9}\)

\(\mathrm{P}\left(\frac{\mathrm{E}_{2}}{\mathrm{A}}\right)=\frac{2}{9}\)

Hence, the correct option is (B).

Given,

A box of oranges.

Let \(A , B\) and \(C\) denotes respectively the events that the first, second and third drawn orange is good.

Now, 

\(P(A)=P\) (good orange in first draw) \(=\frac{12}{15}\)

Because the second orange is drawn without replacement so, now the total number of good oranges will be 11 and the total oranges will be 14 that is the conditional probability of B given that A has already occurred.

Now, 

\(P(\frac{B}{A})=P\) (good orange in second draw) \(=\frac{11}{14}\)

Because the third orange is drawn without replacement so, now the total number of good oranges will be 10 and total orangs will be 13 that is the conditional probability of \(C\) given that \(A\) and \(B\) has already occurred.

Now, 

\(P(\frac{C}{AB})=P\) (good orange in third draw) \(=\frac{10}{13}\)

Thus the probability that all the oranges are good

= \(P(A \cap B \cap C)\) =\(\frac{12}{15} \times \frac{11}{14} \times \frac{10}{13} \)

\(= \frac{44}{91}\)

\(\therefore\) The probability that a box will be approved for sale is \( \frac{44}{91}\).

Hence, the correct option is (A).

Given,

An urn contains 5 red and 5 black balls.

Let in the first attempt the ball drawn is of red colour.

P (probability of drawing a red ball) = \(\frac{5}{10}\) = \(\frac{1}{2}\)

Now the two balls of same colour (red) are added to the urn then the urn contains 7 red and 5 black balls.

P (probability of drawing a red ball) = \(\frac{7}{12}\)

Now,

Let in the first attempt the ball is drawn is of black colour.

P (probability of drawing a black ball) = \(\frac{5}{10}\) = \(\frac{1}{2}\)

Now the two balls of the same colour (black) are added to the urn then the urn contains 5 red and 7 black balls.

P (probability of drawing a red ball) = \(\frac{5}{12}\)

The probability of drawing the second ball as of red colour is \(=\left(\frac{1}{2} \times \frac{7}{12}\right)+\left(\frac{1}{2} \times \frac{5}{12}\right)\)

\(=\frac{1}{2}\left(\frac{7}{12}+\frac{5}{12}\right)=\frac{1}{2} \times 1\)

\(=\frac{1}{2}\)

\(\therefore\) The probability of drawing the second ball as of red colour is \(\frac{1}{2}\).

Hence, the correct option is (D).

Let \(E_{1}\) be the event that the outcome on the die is 5 or \(6, E_{2}\) be the event that the outcome on the die is \(1,2,3\) or 4 and \(\mathrm{A}\) be the event getting exactly head.

Then \(P\left(E_{1}\right)=\frac{2}{6}\)

\(=\frac{1}{ 3}\)

\(P\left(E_{2}\right)=\frac{4}{6}\)

\(=\frac{2}{3}\)

As in throwing a coin three times we get 8 possibilities.

(HHH, HHT, HTH, THH, TTH, THT, HTT, TTT)

\(\Rightarrow P\left(\frac{A}{ E_{1}}\right)=P\) (obtaining exactly one head by tossing the coin three times if she get 5 or 6\()=\frac{3}{8}\)

And \(P\left(\frac{A}{E_{2}}\right)=P\) (obtaining exactly one head by tossing the coin three times if she get \(1,2,3\) or 4\()=\frac{1 }{2}\)

Now the probability that the girl threw \(1,2,3\) or 4 with a die, being given that she obtained exactly one head, is \(\mathrm{P}\) \(\left(\frac{\mathrm{E}_{2}}{\mathrm{A}}\right)\)

By using Bayes' theorem, we have

\(\mathrm{P}\left(\frac{\mathrm{E}_{2}}{ \mathrm{A}}\right)=\frac{\mathrm{P}\left(\mathrm{E}_{2}\right) \cdot \mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{E}_{2}}\right)}{\mathrm{P}\left(\mathrm{E}_{1}\right) \cdot \mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{E}_{1}}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \cdot \mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{E}_{2}}\right)}\)

Now by substituting the values we get

\(=\frac{\frac{2}{3} \cdot \frac{1}{2}}{\frac{1}{3} \cdot \frac{3}{8}+\frac{2}{3} \cdot \frac{1}{2}}\)

\(=\frac{\frac{1}{3}}{\frac{1}{8}+\frac{1}{3}} \)

\(=\frac{\frac{1}{3}}{\frac{3+8}{24}}=\frac{8}{11}\)

\(\Rightarrow \mathrm{P}\left(\frac{\mathrm{E}_{2}}{\mathrm{A}}\right)=\frac{8}{11}\)

Hence, the correct option is (A).

Given,

\(P(A)= 60\%\) = \(\frac{60}{100}\)

\(= 0.60\)

\(P(A') = 1 -0.60\)

\(= 0.40\)

\(P(\frac{B }{A})= 10\%\) = \(\frac{10}{100}\)

\(= 0.10\)

\(P(\frac{B}{A})^{\prime}=80\%\) = \(\frac{80}{100}\)

\(= 0.80\)

We know that,

\(P(\frac{A}{B}) \cdot P(B)=P(\frac{B}{A})-P(A)\)

Let, \(A\) be the event : the employees are graduate

Let, \(B\) be the event : the employees are in sales

\(P(B)=P(A) \cdot P(\frac{B }{ A})+P\left(A^{\prime}\right) \cdot P(\frac{B}{A})^{\prime}\)

\(\Rightarrow P(B) = 0.60 \times 0.10+0.40 \times 0.80\)

\(\Rightarrow P(B) = 0.38\)

\(\therefore\) The probability that an employee selected at random is in sales, is \(0.38\).

Hence, the correct option is (B).

Let us assume, \(X\) denotes the events having a person heart attack

\(\mathrm{A}_{1}\) denote events having the selected person followed the course of yoga and meditation

And, \(\mathrm{A}_{2}\) denote the events having the person adopted the drug prescription

It is given in the question that,

\(P(X)=0.40\)

And, \(P\left(A_{1}\right)=P\left(A_{2}\right)=\frac{1}{2}\)

\(P\left(\frac{X}{A_{1}}\right)=0.40 \times 0.70=0.28\)

\(P\left(\frac{X}{ A_{2}}\right)=0.40 \times 0.75=0.30\)

Therefore,

Probability (The patient suffering from a heart attack and followed a course of meditation and yoga):

\(P\left(\frac{A_{1}}{ X}\right)=\frac{P\left(A_{1}\right) P\left(\frac{X }{ A_{1}}\right)}{P\left(A_{1}\right) P\left(\frac{X}{d A_{1}}\right)+P\left(A_{2}\right) P\left(\frac{X}{A_{2}}\right)}\)

\(=\frac{\frac{1}{2} \times 0.28}{\frac{1}{2} \times 0.28+\frac{1}{2} \times 0.30} \)

\(=\frac{14}{29}\)

Hence, the correct option is (C).

Given,

A lot of 30 bulbs which include 6 defectives.

Then number of non-defective bulbs \(=30-6\)

\(=24\)

As 4 bulbs are drawn at random with replacement.

Let \(X\) denotes the number of defective bulbs from the selected bulbs.

Clearly, \(X\) can take the value of \(0,1,2,3\) or 4 .

\(P(X=0)=P(4\) are non-defective and 0 defective)

\(=4_{C_{0}} \cdot \frac{4}{5}+\frac{4}{5} \cdot \frac{4}{5}.\frac{4}{5}\)

\(=\frac{256}{625}\)

\(P(X=1)=P(3\) are non-defective and 1 defective)

\(=4_{C_{1}} \cdot \frac{1}{5} \cdot\left(\frac{4}{5}\right)^{2}\)

\(=\frac{256}{625}\)

\(P(x=2)=P(2\) are non-defective and 2 defective)

\(=4_{C_{2}} \cdot\left(\frac{1}{5}\right)^{2} \cdot\left(\frac{1}{5}\right)^{2}\)

\(=\frac{96}{625}\)

\(P(X=3)=P(1\) are non-defective and 3 defective)

\(=4_{C_{3}}+\left(\frac{1}{5}\right)^{3}+\frac{4}{5}=\frac{16}{625}\)

\(P(X=4)=P(0\) are non-defective and 4 defective)

\(=4_{C_{4}}.\left(\frac{1}{5}\right)^{4}\)

\(=\frac{1}{625}\)

Hence, the correct option is (C).