Probability Test

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Probability Test - 13

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Question 1
1 / -0

An unbiased coin is tossed. If the outcome is a head then a pair of unbiased dice is rolled and the sum of the numbers obtained on them is noted. If the toss of the coin results in a tail then a card from a well-shuffled pack of nine cards numbered $$1, 2, 3, .., 9$$ is randomly picked and the number on the card is noted. The probability that the noted number is either $$7$$ or $$8$$ is?

A
$$\dfrac{13}{36}$$

B
$$\dfrac{19}{36}$$

C
$$\dfrac{19}{72}$$

D
$$\dfrac{15}{72}$$

Solution

$$P(7\ or\ 8)$$

$$=P(H)P(7\ or\ 8)+P(T)P(7\ or\ 8)$$(head then sum on dice 7 or 8,tell then number on ticket 7 or 8)

$$=\dfrac{1}{2}\times \dfrac{11}{36}+\dfrac{1}{2}\times \dfrac{2}{9}=\dfrac{19}{72}$$.
Question 2
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If $$10$$ different balls are to be placed in $$4$$ distinct boxes at random, then the probability that two of these boxes contain exactly $$2$$ and $$3$$ balls is:

A
$$\dfrac{965}{2^{10}}$$

B
$$\dfrac{945}{2^{10}}$$

C
$$\dfrac{945}{2^{11}}$$

D
$$\dfrac{965}{2^{11}}$$

Solution

$$10$$ different balls
$$4$$ distinct boxes
We can select two boxes in $$^4C_2$$ ways

Total ways in which two boxes will contain exactly two and these balls$$=^4C_2\times ^{10}C_2\times ^8C_2\times 2!\times 2^5$$

Also,
Total ways of distributing $$10$$ different balls in $$4$$ distinct boxes
$$=4^{10}$$

$$\therefore $$ Required probability

$$=\dfrac{^4C_2\times ^{10}C_2\times ^8C_3\times 2^5\times 2!}{4^{10}}$$

$$=\dfrac{\dfrac{4\times 3}{2}\times \dfrac{10\times 9}{2}\times \dfrac{8\times 7\times 12}{6}\times 2^5\times 2}{2^{20}}$$

$$=\dfrac{6\times 45\times 56\times2}{2^{15}}=\dfrac{2\times 3\times 45\times 8\times 7\times 2}{2^{15}}$$

$$=\dfrac{3\times 45\times 7\times 2^5}{2^{15}}=\dfrac{945}{2^{10}}$$
Question 3
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An urn contains $$5$$ red and $$2$$ green balls. A ball is drawn at random from the urn. If the drawn ball is green, then a red ball is added to the urn and if the drawn ball is red, then a green ball is added to the urn; the original ball is not returned to the urn. Now, a second ball is drawn at random from it. The probability that the second ball is red, is :

A
$$\dfrac{26}{49}$$

B
$$\dfrac{32}{49}$$

C
$$\dfrac{27}{49}$$

D
$$\dfrac{21}{49}$$

Solution

$${\textbf{Step -1: Taking the events and framing an equation}}{\text{.}}$$

$${\text{Let }}{E_1}{\text{ be an event of drawing a red ball and placing a green ball in the bag}}$$

  $${E_2}{\text{ be the event of drawing a gree nball and placing a red ball in the bag}}$$

  $${\text{and }}E{\text{ be the event of drawing a red ball in second draw}}{\text{.}}$$

  $${\text{So we can write,}}$$

  $$P(E) = P\left( {{E_1}} \right) \times P\left( {\dfrac{E}{{{E_1}}}} \right) + P\left( {{E_2}} \right) \times P\left( {\dfrac{{E}}{{{E_2}}}} \right)$$

$${\textbf{Step -2: Calculating the required probability}}{\text{.}}$$

  $$P\left( {{E_1}} \right)P\left( {\dfrac{E}{{{E_1}}}} \right) = {\text{drawing a red ball on the second draw when in the first draw we took }}$$

  $${\text{a red ball and placed a green one }}$$

  $$P\left( {{E_1}} \right)P\left( {\dfrac{E}{{{E_1}}}} \right) = \dfrac{5}{7} \times \dfrac{4}{7}$$$${\textbf{ (As earlier there were 5 red balls out of total 7 }}$$

  $${\textbf{and then we drew a red and placed a green so now there are 4 red balls out of 7}}{\text{.)}}$$

  $$P\left( {{E_2}} \right)P\left( {\dfrac{E}{{{E_2}}}} \right) = {\text{drawing a red ball on the second draw when in the first draw we took }}$$

  $${\text{a green ball and placed a red one }}$$

  $$P\left( {{E_2}} \right) \times P\left( {\dfrac{{E}}{{{E_2}}}} \right) = {\text{ }}\dfrac{2}{7} \times \dfrac{6}{7}$$$$\;\;({\textbf{As there were 2 green balls out of 7 and then we put a red ball }}{\text{.}}$$

  $${\textbf{So now we have to draw a red ball and there are 6 reds out of 7}}{\text{.}})$$

  $$P(E) = \dfrac{5}{7} \times \dfrac{4}{7} + \dfrac{2}{7} \times \dfrac{6}{7} = \dfrac{{20 + 12}}{{49}} = \dfrac{{32}}{{49}}$$

$${\textbf{Hence, the total probability is }}\dfrac{{32}}{{49}}.$$
Question 4
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A random variable $$X$$ has the following probability distribution:
$$X$$ $$1$$ $$2$$ $$3$$ $$4$$ $$5$$
$$P(X)$$ $$k^{2}$$ $$2k$$ $$k$$ $$2k$$ $$5k^{2}$$
Then $$P(X>2)$$ is equal to

A
$$\dfrac{1}{6}$$

B
$$\dfrac{7}{12}$$

C
$$\dfrac{1}{36}$$

D
$$\dfrac{23}{36}$$

Solution

$$\displaystyle \sum Pi=1$$

$$\Rightarrow 6k^2+5k=1$$

$$6k^2+5k-1=0$$

$$6k^2+6k-k-1=0$$

$$6k(k+1)-1(k+1)=0$$

$$(k+1)(6k-1)=0$$

$$k=-1,\dfrac{1}{6}$$

$$\therefore\ k=\dfrac 16$$ as k can't be negative

$$p(x>2)=k+2k+5k^2$$

$$=\dfrac{1}{6}+\dfrac{2}{6}+\dfrac{5}{36}$$

$$=\dfrac{23}{36}$$
Question 5
1 / -0

Two different families $$A$$ and $$B$$ are blessed with equal number of children. There are $$3$$ tickets to be distributed amongst the children of these families so that no child gets more than one ticket. If the probability that all the tickets go to the children of the family $$B$$ is $$\displaystyle\dfrac{1}{12}$$, then the number of children in each family is?

A
$$4$$

B
$$6$$

C
$$3$$

D
$$5$$

Solution

There are two different families $$A$$ and $$B$$ with equal number of children.
Let the children in each family be $$x$$.
Thus the total number of children in both the families are $$2x$$

Now, it is given that $$3$$ tickets are distributed amongst the children of the families.
And all the tickets are distributed to the children in family $$B$$.

Thus, the probability that all the three tickets go to the children in family $$B$$ is given by

$$\dfrac{1}{12} = \dfrac{^x C_3}{^{2x} C_3}$$
On solving the above equation, we get,

Thus, $$\dfrac{1}{12} = \dfrac{x(x-1)(x-2)}{2x(2x-1)(2x-2)}$$
Thus, $$\dfrac{1}{3} = \dfrac{x-2}{2x-1}$$

$$\rightarrow 3x-6 = 2x-1$$
$$\rightarrow x = 5$$

Thus, the number of children in each family are $$5$$.
Question 6
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There are $$\mathrm{n}$$ urns each containing $$\mathrm{n}+1$$ balls such that the ith um contains $$\mathrm{i}$$ white balls and $$(\mathrm{n}+1-\mathrm{i})$$ red balls. Let $$\mathrm{u}_{\mathrm{i}}$$ be the event of selecting ith urn, $$\mathrm{i}=1,2,3\ldots,\ \mathrm{n}$$ and $$\mathrm{w}$$ denotes the event of getting a white ball.
If $$\mathrm{n}$$ is even and $$\mathrm{E}$$ denotes the event of choosing even numbered um $$(\displaystyle \mathrm{P}(\mathrm{u}_{\mathrm{i}})=\frac{1}{\mathrm{n}})$$, then the value of $$\mathrm{P}(\mathrm{w}/\mathrm{E})$$ is

A
$$\displaystyle \frac{\mathrm{n}+2}{2\mathrm{n}+1}$$

B
$$\displaystyle \frac{\mathrm{n}+2}{2(\mathrm{n}+1)}$$

C
$$\displaystyle \frac{\mathrm{n}}{\mathrm{n}+1}$$

D
$$\displaystyle \frac{1}{\mathrm{n}+1}$$

Solution

given that $$n$$ is even

$$n(w\cap E )=$$ event of getting white balls from even numbered urn
$$=2+4+6+......+n\\=2(1+2+3+....+\dfrac n2)\\=2\times\dfrac{(n/2)(n/2+1)}{2}\\=\dfrac n4(n+2)$$

$$n(E)=$$ event of selecting even numbered urn
$$=$$ no. of balls in each urn $$\times$$ number of even numbered urns
$$=(n+1)\times \dfrac n2$$

$$\displaystyle { P }\left( \frac { { w } }{ { E } } \right) =\dfrac{n(n+2)/4}{n(n+1)/2}=\dfrac { { n }+2 }{ 2({ n }+1) } $$
Question 7
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A box $$B_{1}$$ contains $$1$$ white ball, $$3$$ red balls and $$2$$ black balls. Another box $$B_{2}$$ contains $$2$$ white balls, $$3$$ red balls and $$4$$ black balls. A third box $$B_{3}$$ contains $$3$$ white balls, $$4$$ red balls and $$5$$ black balls.
If 2 balls are drawn (without replacement) from a randomly selected box and one of the balls is white and the other ball is red, the probability that these 2 balls are drawn from box $$B_{2}$$ is

A
$$\displaystyle \frac{116}{181}$$

B
$$\displaystyle \frac{126}{181}$$

C
$$\displaystyle \frac{65}{181}$$

D
$$\displaystyle \frac{55}{181}$$

Solution

Let A: one ball is white and other is red

$$\mathrm{E}_{1}$$ : both balls are from box $$\mathrm{B}_{1}$$
$$\mathrm{E}_{2}$$ : both balls are from box $$\mathrm{B}_{2}$$
$$\mathrm{E}_{3}$$ : both balls are from box $$\mathrm{B}_{3}$$

Required Probability $$=\displaystyle \mathrm{P}(\frac{\mathrm{E}_{2}}{\mathrm{A}})$$
$$=\frac{\displaystyle
\mathrm{P}(\frac{\mathrm{A}}{\mathrm{E}_{2}})\cdot
\mathrm{P}(\mathrm{E}_{2})}{\displaystyle
\mathrm{P}(\frac{\mathrm{A}}{\mathrm{E}_{1}})\cdot
\mathrm{P}(\mathrm{E}_{1})+\mathrm{P}(\frac{\mathrm{A}}{\mathrm{E}_{2}})\cdot
\mathrm{P}(\mathrm{E}_{2})+\mathrm{P}(\frac{\mathrm{A}}{\mathrm{E}_{3}})\cdot
\mathrm{P}(\mathrm{E}_{3})}$$

$$=\displaystyle
\dfrac{\dfrac{^{2}\mathrm{C}_{1}\times ^{3}\mathrm{C}_{1}}{^{9}\mathrm{C}_{2}}\times\frac{1}{3}}{\dfrac{^{1}\mathrm{C}_{1}\times ^{3}\mathrm{C}_{1}}{^{6}\mathrm{C}_{2}}\times\dfrac{1}{3}+\dfrac{^{2}\mathrm{C}_{1}\times ^{3}\mathrm{C}_{1}}{^{9}\mathrm{C}_{2}}\times\dfrac{1}{3}+\dfrac{^{3}\mathrm{C}_{1}\times ^{4}\mathrm{C}_{1}}{^{12}\mathrm{C}_{2}}\times\dfrac{1}{3}}=\dfrac{\dfrac{1}{6}}{\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{2}{11}}=\dfrac{55}{181}$$
Question 8
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There are $$\mathrm{n}$$ urns each containing $$\mathrm{n}+1$$ balls such that the ith urn contains $$\mathrm{i}$$ white balls and $$(\mathrm{n}+1-\mathrm{i})$$ red balls. Let $$\mathrm{u}_{\mathrm{i}}$$ be the event of selecting ith urn, $$\mathrm{i}=1,2,3\ldots,\ \mathrm{n}$$ and $$\mathrm{w}$$ denotes the event of getting a white ball.

$$\mathrm{I}\mathrm{f}\mathrm{P}(\mathrm{u}_{\mathrm{i}})=\mathrm{c}$$, where $$\mathrm{c}$$ is a constant then $$\mathrm{P}(\mathrm{u}_{\mathrm{n}}/\mathrm{w})$$ is equal to

A
$$\displaystyle \frac{2}{\mathrm{n}+1}$$

B
$$\displaystyle \frac{1}{\mathrm{n}+1}$$

C
$$\displaystyle \frac{\mathrm{n}}{\mathrm{n}+1}$$

D
$$\displaystyle \frac{1}{2}$$

Solution

$$\displaystyle \mathrm{P}(\frac{\mathrm{u}_{\mathrm{n}}}{\mathrm{w}})=\dfrac{\mathrm{c}(\dfrac{\mathrm{n}}{\mathrm{n}+1})}{\mathrm{c}(\dfrac{\Sigma \mathrm{i}}{(\mathrm{n}+1})}$$

$$=\dfrac{\dfrac{n}{n+1}}{\dfrac{n(n+1)}{2(n+1)}}$$

$$=\dfrac{2}{\mathrm{n}+1}$$
Question 9
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One Indian and four American men and their wives are to be seated randomly around a circular table. Then the conditional probability that the Indian man is seated adjacent to his wife given that each American man is seated adjacent to his wife is

A
$$1/2$$

B
$$1/3$$

C
$$2/5$$

D
$$1/5$$

Solution

Let E be the event where an American is seated adjacent to his wife.
A be the event where an Indian is seated adjacent to his wife.

Now
$$n(A\cap E)=4!\times (2!)^{5}$$ (Each couple is taken together as one entity, so we have total 5 things to be arranged in a circle, which can be done in $$(5-1)!=4!$$ number of ways. Also, each couple as one entity can be seated in 2 ways.)

$$n(E)=5!\times(2!)^{4}$$ (Each American couple is taken as one entity, and an Indian man and woman separately. Hence, the arrangement of 6 things in a circle =5! Also each American couple as one entity can be seated in 2 ways. )

$$P\left(\dfrac{A}{E}\right)=\dfrac{n(A\cap E)}{n(E)}=\dfrac{4!\times 32}{5!\times 16}$$

$$=\dfrac{2}{5}$$
Hence required probability is equal to $$\dfrac{2}{5}$$.
Question 10
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A fair die is rolled repeatedly until a six is obtained. Let X denote the number of rolls required.
The conditional probability that $$X \geq 6$$ given $$X > 3$$ equals

A
$$\displaystyle \frac{125}{216}$$

B
$$\displaystyle \frac{25}{216}$$

C
$$\displaystyle \frac{5}{36}$$

D
$$\displaystyle \frac{25}{36}$$

Solution

Let $$X$$ denote the number of rolls required.

For
$$P(X\geq 6)$$

$$=P(6)+P(7)+P(8)...\infty$$

$$=\dfrac{5^{5}}{6^{6}}+\dfrac{5^{6}}{6^{7}}+\dfrac{5^{7}}{6^{8}}+....\infty$$
This is an infinite G.P with common ratio as $$\dfrac{5}{6}$$
Hence
$$Sum=S$$

$$=\dfrac{5^{5}}{6^{6}} \left (\dfrac{6}{6-5} \right )$$

$$=\dfrac{5^{5}}{6^{5}}$$ ...$$(i)$$

For
$$P(X>3)$$

$$P(4)+P(5)+...\infty$$

$$=\dfrac{5^{3}}{6^{4}}+\dfrac{5^{4}}{6^{5}}+\dfrac{5^{5}}{6^{6}}+....\infty$$

$$=\dfrac{5^{3}}{6^{3}}$$ ...$$(ii)$$

Hence applying conditional probability
Equation (i) divide by (ii), we get

$$\text{Required probability}=\dfrac{P(X\ge6)}{P(X>3)}=\dfrac{\dfrac{5^5}{6^5}}{\dfrac{5^3}{6^3}}$$

$$=\dfrac{25}{36}$$

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Re-Attempt Weekly Quiz Competition

$$X$$	$$1$$	$$2$$	$$3$$	$$4$$	$$5$$
$$P(X)$$	$$k^{2}$$	$$2k$$	$$k$$	$$2k$$	$$5k^{2}$$

Probability Test - 13

Result

Probability Test - 13

Score

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Rank

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SHARING IS CARING

Question 1

$$\dfrac{13}{36}$$

$$\dfrac{19}{36}$$

$$\dfrac{19}{72}$$

$$\dfrac{15}{72}$$

Solution

Question 2

$$\dfrac{965}{2^{10}}$$

$$\dfrac{945}{2^{10}}$$

$$\dfrac{945}{2^{11}}$$

$$\dfrac{965}{2^{11}}$$

Solution

Question 3

$$\dfrac{26}{49}$$

$$\dfrac{32}{49}$$

$$\dfrac{27}{49}$$

$$\dfrac{21}{49}$$

Solution

Question 4

$$\dfrac{1}{6}$$

$$\dfrac{7}{12}$$

$$\dfrac{1}{36}$$

$$\dfrac{23}{36}$$

Solution

Question 5

$$4$$

$$6$$

$$3$$

$$5$$

Solution

Question 6

$$\displaystyle \frac{\mathrm{n}+2}{2\mathrm{n}+1}$$

$$\displaystyle \frac{\mathrm{n}+2}{2(\mathrm{n}+1)}$$

$$\displaystyle \frac{\mathrm{n}}{\mathrm{n}+1}$$

$$\displaystyle \frac{1}{\mathrm{n}+1}$$

Solution

Question 7

$$\displaystyle \frac{116}{181}$$

$$\displaystyle \frac{126}{181}$$

$$\displaystyle \frac{65}{181}$$

$$\displaystyle \frac{55}{181}$$

Solution

Question 8

$$\displaystyle \frac{2}{\mathrm{n}+1}$$

$$\displaystyle \frac{1}{\mathrm{n}+1}$$

$$\displaystyle \frac{\mathrm{n}}{\mathrm{n}+1}$$

$$\displaystyle \frac{1}{2}$$

Solution

Question 9

$$1/2$$

$$1/3$$

$$2/5$$

$$1/5$$

Solution

Question 10

$$\displaystyle \frac{125}{216}$$

$$\displaystyle \frac{25}{216}$$

$$\displaystyle \frac{5}{36}$$

$$\displaystyle \frac{25}{36}$$

Solution

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