Question

Choose the correct option.
Radhika and Ridima appear in an interview for two vacancies in the same
organization. The probability of Radhika's selection is (1/5) and of Ridima's selection
is (1/6). What is the probability that only one of them is selected?
A​

Answer 1

The probability that only one of them is selected = 3/10

Given :

• Radhika and Ridima appear in an interview for two vacancies in the same organization.

• The probability of Radhika's selection is 1/5 and of Ridima's selection is 1/6

To find :

The probability that only one of them is selected

Solution :

Step 1 of 2 :

Write down the given probabilities

Here it is given that Radhika and Ridima appear in an interview for two vacancies in the same organization.

The probability of Radhika's selection is 1/5 and of Ridima's selection is 1/6

Let ,

A = The event that Radhika is selected

B = The event that Ridima is selected

Then we have ,

$\displaystyle \sf{P(A) = \frac{1}{5} \: , \: P(B) = \frac{1}{6} }$

Step 2 of 2 :

Find the probability that only one of them is selected

The probability that only one of them is selected

$\displaystyle \sf{ P(A \bar{B} + \bar{A}B )}$

$\displaystyle \sf{ = P(A \bar{B}) + P(\bar{A}B )}$

$\displaystyle \sf{ = P(A) .P(\bar{B}) + P(\bar{A})P(B )}$

$\displaystyle \sf{ = P(A) (1 - P(B)) +(1 - P(A))P(B )}$

$\displaystyle \sf{ = \frac{1}{5} \times \bigg(1 - \frac{1}{6} \bigg) +\frac{1}{6} \times \bigg(1 - \frac{1}{5} \bigg) }$

$\displaystyle \sf{ = \frac{1}{5} \times \bigg( \frac{6 - 1}{6} \bigg) +\frac{1}{6} \times \bigg( \frac{5 - 1}{5} \bigg) }$

$\displaystyle \sf{ = \bigg( \frac{1}{5} \times \frac{5}{6} \bigg) + \bigg( \frac{1}{6} \times \frac{4}{5} \bigg) }$

$\displaystyle \sf{ = \frac{5}{30} + \frac{4}{30} }$

$\displaystyle \sf{ = \frac{5 + 4}{30} }$

$\displaystyle \sf{ = \frac{9}{30} }$

$\displaystyle \sf{ = \frac{3}{10} }$

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