Question

Two students Mehul and Rashi are seeking admission in a college. The probability that Mehul is selected is 0.4 and the probability of selection of exactly one of the them is 0.5. Chances of selection of them is independent of each other. Find the chances of selection of Rashi. Also find the probability of selection of at least one of them.

Answer 1

Let the events be

A: Mehul is selected.

B: Rashi is selected.

Given probability that Mehul is selected = 0.4

P(A) = 0.4

Also, the probability of selection of exactly one of them is 0.5.

P(Mehul selected, but Rashi not selected) + P(Rashi selected, but Mehul not selected) = 0.5

`P(A ∩ barB) + P(B ∩ barA) = 0.5`   ....(1)

Given that events are independent, we can write:

`P(A ∩ barB) = P(A) xx P(barB)`

`P(B ∩ barA) = P(B) xx P(barA)`

Now, the equation (1) becomes:

`P(A ∩ barB) + P(B ∩ barA) = 0.5`

`P(A) xx P(barB) + P(B) xx P(barA) = 0.5`

Using the complement formula:

P(A) × (1 – P(B)) + P(B) × (1 – P(A)) = 0.5

Putting P(A) = 0.4

0.4 × (1 – P(B)) + P(B) × (1 – 0.4) = 0.5

 0.4 – 0.4 × P(B) + P(B) × 0.6 = 0.5

P(B) × 0.6 – P(B) × 0.4 = 0.5 – 0.4

P(B) × (0.6 – 0.4) = 0.1

P(B) × 0.2 = 0.1

`P(B) xx 2/10 = 1/10`

P(B) = `1/10 xx 10/2` 

P(B) = `1/2`

P(B) = 0.5

Thus, the chance of selection of Rashi is 0.5.

We also need to find the probability of selecting at least one of them.

Probability of selection of at least one of them = P(A ∪ B)

= P(A) + P(B) − P(A ∩ B)

Since events are independent,

P(A ∩ B) = P(A) × P(B)

= P(A) + P(B) − P(A) × P(B)

Putting values:

= 0.4 + 0.5 − 0.4 × 0.5

= 0.9 − 0.2

= 0.7