Probability Test - 60

Let us assume \({X}\) 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 \({X}\) has the binomial distribution where \({n}=12\) and \({p}=10 \%=\frac{1}{ 10}\)

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

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

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

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

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

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

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

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

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

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 \(P\left(\frac{A}{E_{2}}\right)\).

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

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

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

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

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

Given, \(P ( A \cup B )\)= \(\frac{5}{6}\) 

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

We know that, \(P(A \cup B)=P(A)+P(B)-P(A \cap B)\)

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

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

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

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

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.

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 \(=P(X=2)\)

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

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

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

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

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

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

\(=0.23\)

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 \(=\mathrm{P}(\mathrm{X} \leq 1)\)

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

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

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

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

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

Given,

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

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 dice

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

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

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

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

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

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

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

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.)

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 \(\frac{5}{16}\).

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 \(=\frac{1}{6}-\frac{25}{216}\)

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

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