Probability Test 23

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Probability Test 23

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Question 1
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An employer sends a letter to his employee but he does not receive the reply (It is certain that employee would have replied if he did receive the letter). It is known that one out of $n$ letters does not reach its destination. Find the probability that employee does not receive the letter.

A
$\displaystyle \frac{1}{n-1}.$

B
$\displaystyle \frac{n}{2n-1}.$

C
$\displaystyle \frac{n-1}{2n-1}.$

D
$\displaystyle \frac{n-2}{n-1}.$

Solution

Let $E$ be the event that employee received the letter and $A$ that employer received the reply, then
$\displaystyle P\left ( E \right )= \frac{n-1}{n}$ and $\displaystyle P\left ( \bar{E} \right )= \frac{1}{n}$
$\displaystyle P\left ( A/E \right )= \frac{n-1}{n}$ and $\displaystyle P\left ( A/\bar{E} \right )= 0$
Now $\displaystyle P\left ( A \right )= P\left ( E\cap A \right )+P\left ( \bar{E}\cap A \right )$
$\displaystyle = P\left ( E \right ).P\left ( A/E \right )+P\left ( \bar{E} \right ).P\left ( A/\bar{E} \right )$
$\displaystyle = \left ( \frac{n-1}{n} \right )\left ( \frac{n-1}{n} \right )+\frac{1}{n}.0$
$\displaystyle P\left ( A \right )= \left ( \frac{n-1}{n} \right )^{2}$
$$\displaystyle
P\left ( \bar{A} \right )= 1-\left ( \frac{n-1}{n} \right )^{2}=
\frac{n^{2}-n^{2}-1+2n}{n^{2}}= \frac{2n-1}{n^{2}}$$
Now the required probability
$$\displaystyle P\left ( E/\bar{A} \right )= \frac{P\left ( E\cap \bar{A} \right )}{P\left ( \bar{A} \right )}= \frac{P\left ( E \right )-P\left ( E\cap A

\right )}{P\left ( \bar{A} \right )}$$

$\displaystyle = \frac{P\left ( E \right )-P\left ( E \right ).P\left ( A/E \right )}{P\left ( \bar{A} \right )}$
Putting the values, we get
$\displaystyle = \dfrac{\dfrac{n-1}{n}-\dfrac{n-1}{n}.\dfrac{n-1}{n}}{\dfrac{2n-1}{n^{2}}}$
$\displaystyle \therefore P\left ( E/\bar{A} \right )= \frac{n-1}{2n-1}.$
Question 2
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Directions For Questions

A class consists of $n$ students. For $0\leq k\leq n$ , let $E_k$ denote the event that exactly $k$ student out of $n$ pass in the examination. Let $P(E_k)=p_k$ and let $A$ denote the event that a student $X$ selected at random pass in the examination.

...view full instructions

If $P(E_k)\propto k$ for $0\leq k\leq n$ , then the probability that X is the only student to pass the examination is

A
$\displaystyle \frac {3}{n(n+1)}$

B
$\displaystyle \frac {6}{n(n+1)(2n+1)}$

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

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

Solution

$\displaystyle \sum_{k = 0}^{k = n}P(E_k) = 1$
$\therefore \displaystyle \sum_{k = 0}^{k = n}\alpha k = 1$
$\therefore \alpha = \dfrac{2}{n(n + 1)}$
Now,
$P(A) = \displaystyle \sum_{k = 0}^{k = n}P(E_k)\times P(A/E_k)$
$P(A/E_k) = \dfrac{k}{n}$
$\therefore P(A) = \dfrac{2}{n^2(n + 1)}\displaystyle \sum_{k = 0}^{k = n}k^2$
$\therefore P(A) = \dfrac{2}{n^2(n + 1)}\times \dfrac{n(n + 1)(2n + 1)}{6} = \dfrac{2n+1}{3n}$
Probability that only $X$ passes $=P(E_1/A) = \dfrac{P(E_1)P(A/E_1)}{P(A)}$
$=\dfrac{2}{n(n + 1)}\dfrac{1}{n}\times\dfrac{3n}{(2n+1)}$
$=\dfrac{6}{n(n+1)(2n+1)}$
Question 3
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Directions For Questions

A class consists of $n$ students. For $0\leq k\leq n$ , let $E_k$ denote the event that exactly $k$ student out of $n$ pass in the examination. Let $P(E_k)=p_k$ and let $A$ denote the event that a student $X$ selected at random pass in the examination.

...view full instructions

If $P(E_k)=C$ for $0\leq k\leq n$ , then $P(A)$ equals

A
$1/2$

B
$2/3$

C
$1/6$

D
$1/(n+1)$

Solution

$P(E_k)=C$ for $0\leq k\leq n$ and $\displaystyle \sum_{k=0}^nP(E_k)=1\Rightarrow C=1/(n+1)$ .
Also $\displaystyle P(A|E_k)=\frac {^{n-1}C_{k-1}}{^nC_k}=\frac {k}{n}$
By the total probability rule
$P(A)=\displaystyle \sum_{k=0}^nP(E_k)P(A|E_k)$
$\displaystyle =\frac {1}{n(n+1)}\frac {n(n+1)}{2}=\frac {1}{2}$ .
Question 4
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Directions For Questions

A class consists of $n$ students. For $0\leq k\leq n$ , let $E_k$ denote the event that exactly $k$ student out of $n$ pass in the examination. Let $P(E_k)=p_k$ and let $A$ denote the event that a student $X$ selected at random pass in the examination.

...view full instructions

If $P(E_k)\propto k$ for $0\leq k\leq n$ , then $\displaystyle \lim_{n\rightarrow \infty}\sum_{k=0}^nP(E_k|A)$ equals

A
$0$

B
$1/2$

C
$1/6$

D
$1$

Solution

We have by Baye's theorem
$P(E_k|A)=\cfrac{P(E_k)P(A|E_k)}{P(A)}$ ....(1)

Given $P(E_k)\propto k$ for $0 \le k \le n$
$\Rightarrow P(E_k)=p_k =kk_1$ where $k_1$ is constant of proportionality.

Since, $\sum_{k=0}^n P(E_k) =1$
$\Rightarrow p_0+ p_1+p_2+ .....+p_n=1$
$\Rightarrow 0.k_1+k_1+2k_1+ 3k_1+.....nk_1=1$
$\Rightarrow k_1 (1+2+3+....+n)=1$
$\Rightarrow k_1=\displaystyle \frac{2}{n(n+1)}$
$\Rightarrow P(E_k)=p_k=\cfrac{2k}{n(n+1)}$

Now, $P(A|E_k)=\cfrac{P(A\cap E_k)}{P(E_k)}$
$\Rightarrow P(A|E_k)=\cfrac{^{n-1}C_{k-1}}{^nC_k}$
$\Rightarrow P(A|E_k)=\cfrac{(n-1)!/(n-k)!(k-1)!}{n!/(n-k)!k!}$
$\Rightarrow P(A|E_k)=\cfrac{k}{n}$

Now, $P(A)=\sum_{k=0}^{n}P(E_k)P(A|E_k)$
$P(A)=0+\cfrac{2}{n(n+1)} \cfrac{1}{n}+\cfrac{2.2}{n(n+1)}\cfrac{2}{n}+\cfrac{2.3}{n(n+1)}\cfrac{3}{n}+......\cfrac{2.n}{n(n+1)}\cfrac{n}{n}$

$\Rightarrow P(A)=\cfrac{2}{n^2(n+1)} (1^2+2^2+3^2+....+n^2)$
$\Rightarrow P(A)=\cfrac{2}{n^2(n+1)} \cfrac{n(n+1)(2n+1)}{6}$
$\Rightarrow P(A)=\cfrac{2n+1}{3n}$

Substituting these values in (1), we get
$P(E_{ k }|A)=\cfrac { \cfrac { 2k }{ n(n+1) } \cfrac { k }{ n } }{ \cfrac { 2n+1 }{ 3n } }$
$P(E_k|A)=\cfrac{6k^2}{n(n+1)(2n+1)}$
Question 5
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A box contain $N$ coins, $m$ of which are fair are rest and biased. The probability of getting a head when a fair coin is tossed is $1/2$ , while it is $2/3$ when a biased coin is tossed. A coin is drawn from the box at random and is tossed twice. The first time it shows head and the second time it shows tail. The probability that the coin drawn is fair is

A
$\dfrac {9m}{8N+m}$

B
$\dfrac {m}{8N+m}$

C
$\dfrac {N}{8n+m}$

D
$\dfrac {1}{m}$

Solution

Let $E_1, E_2$ and $A$ denote the following events:
$E_1$ : coin selected is fair
$E_2$ : coin selected is biased
$A:$ the first toss results in a head and the second toss results in a tail.
$\displaystyle P(E_1)=\frac {m}{N}, P(E_2)=\frac {N-m}{N}$ ,
$\displaystyle P(A|E_1)=\frac {1}{2}\times \frac {1}{2}\times \frac {1}{4}, P(A|E_2)=\frac {2}{3}\times \frac {1}{3}=\frac {2}{9}$ .
By Bayes' rule
$\displaystyle P(E_1|A)=\frac {P(E_1)P(A|E_1)}{P(E_1)P(A|E_1)+P(E_2)P(A|E_2)}=\frac {9m}{8N+m}$ .
Question 6
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Directions For Questions

A class consists of $n$ students. For $0\leq k\leq n$ , let $E_k$ denote the event that exactly $k$ student out of $n$ pass in the examination. Let $P(E_k)=p_k$ and let $A$ denote the event that a student $X$ selected at random pass in the examination.

...view full instructions

If $P(E_k)\propto k$ for $0\leq k\leq n$ , the P(A) equals

A
$3n/(4n+1)$

B
$(2n+1)/3n$

C
$1/(n+1)$

D
$1/n^2$

Solution

Let $P(E_k)=k\alpha$ where $\alpha$ is the constant of proportionality.
Since $\displaystyle \sum_{k=0}^nP(E_k)=1$ , we get $\alpha=2/n(n+1)$
Now, $P(A)=\displaystyle \sum_{k=0}^nP(E_k)P(A|E_k)=\frac {2}{n^2(n+1)}\sum_{k=0}^nk^2$
$\displaystyle =\frac {2}{n^2(n+1)}\frac {n(n+1)(2n+1)}{6}=\frac {2n+1}{3n}$
Question 7
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Twelve players ${ S }_{ 1 },{ S }_{ 2 },...,{ S }_{ 12 }$ play in a chess tournament. They are divided into six pairs at random. From each pair a winner is decided. It is assumed that all players are of equal strength. The probability that exactly one of ${S}_{1}$ and ${S}_{2}$ is among the six winners is

A
$\displaystyle \frac { 6 }{ 11 }$

B
$\displaystyle \frac { 5 }{ 11 }$

C
$\displaystyle \frac { 4 }{ 11 }$

D
None of these

Solution

Let $B=$ {exactly one of ${S}_{1}$ and ${S}_{2}$ is among of the six winners}.
Case I. $({S}_{1}$ and ${S}_{2}$ are paired)
Let ${E}_{1}=$ { ${S}_{1}$ and ${S}_{2}$ are paired}.
Since ${S}_{1}$ can be paired with any of the remaining $11$ players, therefore $\displaystyle P\left( { E }_{ 1 } \right) =\frac { 1 }{ 11 } .$
In this case, it is certain that exactly one of ${S}_{1}$ and ${S}_{2}$ will winner, so that $\displaystyle P\left( \frac { B }{ { E }_{ 1 } } \right) =1.$

Case II. ( ${S}_{1}$ and ${S}_{2}$ are not paired)
Let ${E}_{2}=$ { ${S}_{1}$ and ${S}_{2}$ are not paired}, then $\displaystyle P\left( { E }_{ 2 } \right) =\frac { 10 }{ 11 } .$
In this case, we have
$\displaystyle P\left( \frac { B }{ { E }_{ 2 } } \right) =P\left( { S }_{ 1 }\cap { S' }_{ 2 } \right) +P\left( { S }'_{ 1 }\cap { S }_{ 2 } \right)$
$\displaystyle =P\left( { S }_{ 1 } \right) P\left( { S' }_{ 2 } \right) +P\left( { S' }_{ 1 } \right) P\left( { S }_{ 2 } \right)$
$\displaystyle =\frac { 1 }{ 2 } .\frac { 1 }{ 2 } +\frac { 1 }{ 2 } .\frac { 1 }{ 2 } =\frac { 1 }{ 2 }$
[Since all players are of equal strength, therefore, $\displaystyle P({S}_{1})=\frac { 1 }{ 2 }$ and $\displaystyle P\left( { S' }_{ 1 } \right) =\frac { 1 }{ 2 } ]$
$\displaystyle \therefore P\left( B \right) =P\left( { E }_{ 1 } \right) P\left( \frac { B }{ { E }_{ 1 } } \right) +P\left( { E }_{ 2 } \right) P\left( \frac { B }{ { E }_{ 2 } } \right)$
$\displaystyle =\frac { 1 }{ 11 } .1+\frac { 10 }{ 11 } .\frac { 1 }{ 2 } =\frac { 6 }{ 11 } .$
Question 8
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A bag contains $(2n+1)$ coins. It is known that $n$ of these coins have a head on both sides, whereas the remaining $n+1$ coins are fair. A coin is picked up at random from the bag and tossed. If the probability that the toss results in a head is $\displaystyle \frac{31}{42}$ , then $n$ is equal to

A
$10$

B
$11$

C
$12$

D
$13$

Solution

Let ${ A }_{ 1 }$ denote the event that a coin having heads on both sides is chosen, and ${ A }_{ 2 }$ denote the vent that a fiar coin is chosen.
Let $E$ denote the vent that head occurs, Then
$\displaystyle P\left( { A }_{ 1 } \right) =\frac { n }{ 2n+1 } \Rightarrow P\left( { A }_{ 2 } \right) =\frac { n+1 }{ 2n+1 }$
Probability of occurrence of event $E$ , if unfair coin was selected is $\displaystyle P\left( \frac { E }{ { A }_{ 1 } } \right) =1$
Probability of occurrence of event $E$ , if fair coin was selected is $\displaystyle P\left( \frac { E }{ { A }_{ 2 } } \right) =\frac { 1 }{ 2 }$
$\because P\left( E \right) =P\left( { A }_{ 1 }\cap E \right) +P\left( { A }_{ 2 }\cap E \right)$
$\displaystyle \therefore P\left( E \right) =P\left( { A }_{ 1 } \right) P\left( \frac { E }{ { A }_{ 1 } } \right) +P\left( { A }_{ 2 } \right) P\left( \frac { E }{ { A }_{ 2 } } \right)$
$\displaystyle \Rightarrow \frac { 31 }{ 42 } =\frac { n }{ 2n+1 } .1+\frac { n+1 }{ 2n+1 } .\frac { 1 }{ 2 } \Rightarrow \frac { 31 }{ 42 } =\frac { 3n+1 }{ 2\left( 2n+1 \right) } \\ \Rightarrow 124n+62=126n+42\Rightarrow 2n=20\Rightarrow n=10$
Question 9
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Directions For Questions

A JEE aspirant estimates that she will be successful with an $80$ percent chance if she studies $10$ hours per day, with a $60$ percent chance if she studies 7 hours per day and with a 40 percent chance if she studies 4 hours per day. She further believes that she will study 10 hours, 7 hours and 4 hours per day with probabilities 0.1, 0.2 and 0.7, respectively.

...view full instructions

Given that she is successful, the chance she studied for 4 hours, is

A
$\displaystyle \frac {6}{12}$

B
$\displaystyle \frac {7}{12}$

C
$\displaystyle \frac {8}{12}$

D
$\displaystyle \frac {9}{12}$

Solution

Given,The probability of a Jee Aspirant to be successful if he studies for 10 hours per day, $P(t_{10})=0.8$
The probability of a Jee Aspirant to be successful if he studies for 7 hours per day, $P(t_{7})=0.6$
The probability of a Jee Aspirant to be successful if he studies for 4 hours per day, $P(t_{4})=0.4$
The probability that she will study 10 hours per day, $\displaystyle P(A)=0.1$
The probability that she will study 7 hours per day, $\displaystyle P (B)=0.2$
The probability that she will study 4 hours per day, $\displaystyle P(C)=0.7$
$\therefore The\;probability\;that\;she\;will\;be\;successful,P(S)=P(t_{10})*P(A)+P(t_{7})*P(B)+P(t_{4})*P(S/t_{4})$
$=0.8*0.1+0.6*0.2+0.4*0.7=0.48$
But probability that she studies 4 hours per day for success= $P(t_{4} \cap C)=P(t_{4})P(C)=0.4*0.7=0.28$
Since $(A,t_{10}),(B,t_{7}),(C,t_{4})$ pairwise independent $\Rightarrow P(A \cap B)=P(A).P(B)$
$\therefore$ The probability that she studies 4 hours per day given she is successful, $P(C/S)=\displaystyle\frac{P(C \cap S)}{P(S)}$
$=\displaystyle\frac{0.28}{0.48}=\displaystyle\frac{7}{12}$
Question 10
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The contents of urn I and II are as follows:
Urn I: 4 white and 5 black balls
Urn II: 3 white and 6 black balls
One urn is chosen at random and a ball is drawn and its colour is noted and replaced back to the urn. Again a ball is drawn from the same urn colour is noted and replaced. The process is repeated 4 times and as a result one ball of white colour and 3 of black colour are noted. Find the probability the chosen urn was I.

A
$\displaystyle \frac{125}{287}$

B
$\displaystyle \frac{64}{127}$

C
$\displaystyle \frac{25}{287}$

D
$\displaystyle \frac{79}{192}$

Solution

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Probability Test 23

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Probability Test 23

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Question 1

1n−1.\displaystyle \frac{1}{n-1}.n−11​.

n2n−1.\displaystyle \frac{n}{2n-1}.2n−1n​.

n−12n−1.\displaystyle \frac{n-1}{2n-1}.2n−1n−1​.

n−2n−1.\displaystyle \frac{n-2}{n-1}.n−1n−2​.

Solution

Question 2

3n(n+1)\displaystyle \frac {3}{n(n+1)}n(n+1)3​

6n(n+1)(2n+1)\displaystyle \frac {6}{n(n+1)(2n+1)}n(n+1)(2n+1)6​

1n\displaystyle \frac {1}{n}n1​

1n(2n+1)\displaystyle \frac {1}{n(2n+1)}n(2n+1)1​

Solution

Question 3

1/21/21/2

2/32/32/3

1/61/61/6

1/(n+1)1/(n+1)1/(n+1)

Solution

Question 4

000

1/21/21/2

1/61/61/6

111

Solution

Question 5

9m8N+m\dfrac {9m}{8N+m}8N+m9m​

m8N+m\dfrac {m}{8N+m}8N+mm​

N8n+m\dfrac {N}{8n+m}8n+mN​

1m\dfrac {1}{m}m1​

Solution

Question 6

3n/(4n+1)3n/(4n+1)3n/(4n+1)

(2n+1)/3n(2n+1)/3n(2n+1)/3n

1/(n+1)1/(n+1)1/(n+1)

1/n21/n^21/n2

Solution

Question 7

611\displaystyle \frac { 6 }{ 11 } 116​

511\displaystyle \frac { 5 }{ 11 } 115​

411\displaystyle \frac { 4 }{ 11 } 114​

None of these

Solution

Question 8

101010

111111

121212

131313

Solution

Question 9

612\displaystyle \frac {6}{12}126​

712\displaystyle \frac {7}{12}127​

812\displaystyle \frac {8}{12}128​

912\displaystyle \frac {9}{12}129​

Solution

Question 10

125287\displaystyle \frac{125}{287}287125​

64127\displaystyle \frac{64}{127}12764​

25287\displaystyle \frac{25}{287}28725​

79192\displaystyle \frac{79}{192}19279​

Solution

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$\displaystyle \frac{1}{n-1}.$

$\displaystyle \frac{n}{2n-1}.$

$\displaystyle \frac{n-1}{2n-1}.$

$\displaystyle \frac{n-2}{n-1}.$

$\displaystyle \frac {3}{n(n+1)}$

$\displaystyle \frac {6}{n(n+1)(2n+1)}$

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

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

$1/2$

$2/3$

$1/6$

$1/(n+1)$

$0$

$1/2$

$1/6$

$1$

$\dfrac {9m}{8N+m}$

$\dfrac {m}{8N+m}$

$\dfrac {N}{8n+m}$

$\dfrac {1}{m}$

$3n/(4n+1)$

$(2n+1)/3n$

$1/(n+1)$

$1/n^2$

$\displaystyle \frac { 6 }{ 11 }$

$\displaystyle \frac { 5 }{ 11 }$

$\displaystyle \frac { 4 }{ 11 }$

$10$

$11$

$12$

$13$

$\displaystyle \frac {6}{12}$

$\displaystyle \frac {7}{12}$

$\displaystyle \frac {8}{12}$

$\displaystyle \frac {9}{12}$

$\displaystyle \frac{125}{287}$

$\displaystyle \frac{64}{127}$

$\displaystyle \frac{25}{287}$

$\displaystyle \frac{79}{192}$