GATE CSE 2017 Set 2 | Question: 26

Question

$P$ 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

• A. $\left(\dfrac{4}{5}\right)$
• B. $\left(\dfrac{5}{6}\right)$
• C. $\left(\dfrac{7}{8}\right)$
• D. $\left(\dfrac{11}{12}\right)$

Answer 1

Let,

$P =$ P applies for the job

$Q =$ Q applies for the job

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$P(P) = \frac{1}{4} \to(1)$  

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

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

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: $A$

Comments

you can also see the fact that they are dependent by seeing the outcome of one if given another ,the outcomes are changing because of another so we say events here are dependent .

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can anyone tell why we can't write like this  (P¯ ⋃ Q¯)= 1 − P(P ⋂ Q)

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You can. But how does it help in this question?

Answer 2

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

Comments

P(p'/q)=1-1/6 =5/6 , should be 1/2?

P(p'/q)= 1-p(p/q) so 1-1/2= 1/2 ??

 

Answer is anyways correct.

Answer 3

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 4

hope it might help........