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. $\left(\dfrac{11}{12}\right)$

Comments

sol

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Ans

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Note : This can be Also solve using "Tree Method" if u consider P( Q | P' ) = x and P(Q' | P' ) = 1-x 

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$$
P(P)=\frac14,\qquad P(P^c)=1-P(P)=\frac34
$$

$$
P(P\mid Q)=\frac12,\qquad P(Q\mid P)=\frac13
$$

$$
P(P\cap Q)=P(Q\mid P)P(P)
$$

$$
=\frac13\cdot\frac14=\frac1{12}
$$

$$
P(P\mid Q)=\frac{P(P\cap Q)}{P(Q)}
$$

$$
\frac12=\frac{\frac1{12}}{P(Q)}
$$

$$
P(Q)=\frac{\frac1{12}}{\frac12}=\frac16
$$

$$
P(Q^c)=1-P(Q)
$$

$$
=1-\frac16=\frac56
$$

$$
P(P\cap Q^c)=P(P)-P(P\cap Q)
$$

$$
=\frac14-\frac1{12}
=\frac3{12}-\frac1{12}
=\frac2{12}
=\frac16
$$

$$
P(P\mid Q^c)
=\frac{P(P\cap Q^c)}{P(Q^c)}
$$

$$
=\frac{\frac16}{\frac56}
=\frac15
$$

$$
P(P^c\mid Q^c)=1-P(P\mid Q^c)
$$

$$
=1-\frac15
=\boxed{\frac45}
$$

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

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?

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@gurichouhan Because by De Morgan's law $1-(P\cap Q)=(P\cap Q)'=P'\cup Q'$

Answer 2

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 3

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 4

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

Answer 5

@GO Classes 

Comments

@Snehashish007 How did you come to know that 2 trees would be formed? What was the intuition behind it?

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See, our goal is to find P(p’|q’) which makes tree2 obvious. In tree2 I had to find x also I had P(Q|P) and P(P) from which I found P(P intersection Qcomp.) for which I drew tree1.

Note: Sachin Sir has solved many more complicated questions using tree method so getting the intuition was not hard.

Answer 6

Probability that ‘P’ applies for a job = 1/4 = P(p) ⇾ (1)
Probability that ‘P’ applies for the job given that Q applies for the job = P(p/q) = 1/2 ⇾ (2)
Probability that ‘Q’ applies for the job, given that ‘P’ applies for the job = P(p/q) = 1/3 ⇾ (3)
Bayes Theorem:
(P(A/B) = (P(B/A)∙P(A))/P(B) ; P(A/B) = P(A∩B)/P(B))
⇒ P(p/q) = (P(q/p)∙P(p))/p(q)
⇒ 1/2 = (1/3×1/4)/p(q)
p(q) = 1/12×2 = 1/(6) (P(q) = 1/6) ⇾ (4)
From Bayes,
P(p/q) = (P(p∩q))/(P(q))
1/2 = P(p∩q)/(1⁄6)
(p(p∩q) = 1/12)
We need to find out the “probability that ‘P’ does not apply for the job given that q does not apply for the job = P(p'/q')
From Bayes theorem,
P(p'/q') = (P(p'∩q'))/P(q') ⇾ (5)
We know,
p(A∩B) = P(A) + P(B) - P(A∪B)
also (P(A'∩B') = 1 - P(A∪B))
P(p'∩q') = 1 - P(p∪q)
= 1 - (P(p) + P(q) - P(p∩q))
= 1 - (P(p) + P(q) - P(p) ∙ P(q))
= 1 - (1/4 + 1/6 - 1/12)
= 1 - (10/24 - 2/24)
= 1 - (8/24)
= 2/3
(P(p'∩q') = 2/3) ⇾ (6)
Substitute in (5),
P(p'⁄q') = (2⁄3)/(1-P(q)) = (2⁄3)/(1-1/6) = (2⁄3)/(5⁄6) = 4/5
(P(p'/q') = 4/5)

OPTION: A