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

Answer 2

Solve it using tree diagram...It is very simple

Answer 3

Ans

Answer 4

 

Conditional Probability Formula:

The probability of an event A occurring given that event B has already occurred is:

P(A | B) = P(A ∩ B) / P(B)

De Morgan's Law:

To find the probability of neither event occurring, we look at the complement of at least one event occurring:

P(P' ∩ Q') = P((P ∪ Q)') = 1 - P(P ∪ Q)

Given Data:

1. P(P) = 1/4 (Probability P applies)

2. P(P | Q) = 1/2 (Probability P applies given Q applies)

3. P(Q | P) = 1/3 (Probability Q applies given P applies)

Goal:

Find P(P' | Q') (Probability P does NOT apply given Q does NOT apply).

Formula:

P(P' | Q') = P(P' ∩ Q') / P(Q')

Find the Intersection P(P ∩ Q)

Using the conditional formula for P(Q | P):

P(Q | P) = P(P ∩ Q) / P(P)

1/3 = P(P ∩ Q) / (1/4)

⇒ P(P ∩ Q) = (1/3) × (1/4) = 1/12

Find the Total Probability of Q, P(Q)

Using the conditional formula for P(P | Q):

P(P | Q) = P(P ∩ Q) / P(Q)

1/2 = (1/12) / P(Q)

⇒ P(Q) = (1/12) × 2 = 1/6

Find the Denominator P(Q')

P(Q') = 1 - P(Q)

P(Q') = 1 - 1/6 = 5/6

Find the Numerator P(P' ∩ Q') Using De Morgan's Law

First find the total union area:

P(P ∪ Q) = P(P) + P(Q) - P(P ∩ Q)

P(P ∪ Q) = 1/4 + 1/6 - 1/12

Making denominators common (LCM = 12):

P(P ∪ Q) = 3/12 + 2/12 - 1/12

P(P ∪ Q) = 4/12 = 1/3

Now apply De Morgan's Law:

P(P' ∩ Q') = 1 - P(P ∪ Q)

P(P' ∩ Q') = 1 - 1/3 = 2/3

Final Division

P(P' | Q') = P(P' ∩ Q') / P(Q')

P(P' | Q') = (2/3) / (5/6)

P(P' | Q') = (2/3) × (6/5)

P(P' | Q') = 12/15 = 4/5