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$P$ and $Q$ are considering to apply for a job. The probability that $P$ applies for the job is $\dfrac{1}{4},$ the probability that $P$ applies for the job given that $Q$ applies for the job is $\dfrac{1}{2},$ and the probability that $Q$ applies for the job given that $P$ applies for the job is $\dfrac{1}{3}.$ Then the probability that $P$ does not apply for the job given that $Q$ does not apply for this job is

  1. $\left(\dfrac{4}{5}\right)$
  2. $\left(\dfrac{5}{6}\right)$
  3. $\left(\dfrac{7}{8}\right)$
  4. $\left(\dfrac{11}{12}\right)$
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Best answer
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86 votes

Let,

$P =$ P applies for the job

$Q =$ Q applies for the job


$P(P) = \frac{1}{4} \to(1)$  

$P\left({P}\mid{Q}\right) = \frac{1}{2} \to(2)$

$P\left({Q}\mid{P}\right) = \frac{1}{3} \to (3)$

Now, we need to find $P\left({P'}\mid{Q'}\right)$ 

From $(2)$

$P({P}\mid {Q}) =\frac{P(P\cap Q)}{P(Q)} = \frac{1}{2}\to (4)$

From $(1)$ and $(3),$

$P({Q}\mid {P}) = \frac{P(P\cap Q)}{P(P)} = \frac{P(P\cap Q)}{\frac{1}{4}} = \frac{1}{3}$
$\therefore$ $P(P\cap Q) = \frac{1}{12}\to (5)$ 

From $(4)$ and $(5),$

$P(Q) = \frac{1}{6} \to (6)$

Now, $P({P'}\mid {Q'}) = \frac{P(P' \cap Q')}{P(Q')} \to (7)$

From $(6)$
$P(Q') = 1 - 1/6 = 5/6 \to (8)$

Also, $P(P' \cap Q') = 1 - [ P(P \cup Q) ]$

$\quad = 1 - [ P(P) + P(Q) - P(P\cap Q) ]$
$\quad = 1 - [ 1/4 + 1/6 - 1/12 ]$
$\quad = 1 - [ 1/3 ]$
$\quad = 2/3 \to (9)$

Hence, from $(7),(8)$ and $(9)$

$P(P' \mid Q')= \frac{\frac{2}{3}}{\frac{5}{6}} = \frac{4}{5}.$

Correct Answer: $A$

edited by
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15 votes
Ans  A 4/5

P(p)=1/4,   P(p/q)=1/2 ,P(q/p)=1/3

by Bayes theorem

P(p/q)=P(q/p).P(p) /P(q) =>1/2=((1/3)*1/4)/P(q)=>P(q)=1/6 and P(q')=5/6

P(q'/p)=2/3

P(p/q')=P(q'/p).P(p) / P(q') =(2/3*1/4)  / 5/6 =>1/5

 

P(p'/q')=1-P(p/q') =1-1/5=4/5

correct me if i am wrong
edited by
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13 votes

Solve by Tree Diagram:

Given that, P($P$ $apply$ $for$ $job$) = $\frac{1}{4}$

P($P$ $apply$ $for$ $job$ | $Q$ $apply$ $for$ $job$ ) = $\frac{1}{2}$

Let, the probability of $Q$ does not apply for job given that $P$ does not apply for job = $x$

Probability of Q apply for job = $\frac{1}{4} \times \frac{1}{3} + \frac{3}{4} \times (1 – x)$

$\therefore$  P($P$ $apply$ $for$ $job$ | $Q$ $apply$ $for$ $job$ ) = $\frac{1}{2}$

$=>$ $\frac{P(P \cap Q)}{P(Q)}$ = $\frac{1}{2}$ [Here, P = P apply for job; Q = Q apply for job]

$=>$ $\frac{\frac{1}{4} \times \frac{1}{3}}{\frac{1}{4} \times \frac{1}{3} + \frac{3}{4} \times (1 – x)}$ = $\frac{1}{2}$

$=>$ $x$ = $\frac{8}{9}$

Probability of Q NOT apply for job = $\frac{1}{4} \times \frac{2}{3} + \frac{3}{4} \times \frac{8}{9}$

$\therefore$ The probability of $P$ does not apply for job given that $Q$ does not apply for job:

P($P$ $NOT$ $apply$ $for$ $job$ | $Q$ $NOT$ $apply$ $for$ $job$):

$\frac{P(P^{c} \cap Q^{c})}{P(Q^{c})}$ = $\frac{\frac{3}{4} \times \frac{8}{9}}{\frac{1}{4} \times \frac{2}{3} + \frac{3}{4} \times \frac{8}{9}}$ = $\frac{4}{5}$ 

[Ans]: A.

Answer:

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