Probability Test - 61

The sample space of the experiment is,

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

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,

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

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

\(=0\)

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.

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 eventsthat 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.

\(=P(A \cap B)=\frac{1}{2} \times \frac{25}{51}=\frac{25}{102}\)

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

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 )

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

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

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

\(=1-\mathrm{p}(0)\) \(=1-\left(\frac{9}{10}\right)^{10}\)

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

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

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

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,

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

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

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

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

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

Given,

A dice 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 \(\frac{7}{8}\).

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) =\(\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\)

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

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

Also, \({X}\) 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 \(=P(X \geq 4)\)

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

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

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

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

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

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 \(\mathrm{II}\)

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

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

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

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

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