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In the MBA Programme of a B - School, there are two sections A and B. 1/4th of the students in Section A and 4/9th of the students in section B are girls. If two students are chosen at random, one each from section A and Section B as class representative, the probability that exactly one of the students chosen is a girl, is :
  • a)
    23/72
  • b)
    11/36
  • c)
    5/12
  • d)
    17/36
Correct answer is option 'D'. Can you explain this answer?
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In the MBA Programme of a B - School, there are two sections A and B. ...
Probability of choosing a boy from Section A = 3/4
Probability of choosing a boy from Section B = 5/9
Probability of choosing a girl from Section A = 1/4
Probability of choosing a girl from Section B = 4/9
There are two cases. 
Case 1: Boy from section A and girl from section B
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In the MBA Programme of a B - School, there are two sections A and B. ...
Problem:
In the MBA Programme of a B-School, there are two sections A and B. 1/4th of the students in Section A and 4/9th of the students in Section B are girls. If two students are chosen at random, one each from section A and section B as class representatives, the probability that exactly one of the students chosen is a girl is:

a) 23/72
b) 11/36
c) 5/12
d) 17/36

Solution:
To find the probability that exactly one of the students chosen is a girl, we need to consider the cases where one student is a girl and the other is a boy from both sections A and B.

Let's assume that there are a total of 'x' students in section A and 'y' students in section B.

Given that 1/4th of the students in section A are girls, the number of girls in section A can be calculated as (1/4)x. Similarly, 4/9th of the students in section B are girls, so the number of girls in section B can be calculated as (4/9)y.

Now, let's consider the cases where one student is a girl and the other is a boy:

1. The first student chosen is a girl from section A and the second student chosen is a boy from section B. The probability of this happening can be calculated as:
(1/4)x * (5/9)y / (x * y)

2. The first student chosen is a boy from section A and the second student chosen is a girl from section B. The probability of this happening can be calculated as:
(3/4)x * (4/9)y / (x * y)

Adding these probabilities, we get the total probability of exactly one student chosen being a girl as:

[(1/4)x * (5/9)y / (x * y)] + [(3/4)x * (4/9)y / (x * y)]

Simplifying this expression, we get:

(5/36) + (12/36)

= 17/36

Therefore, the correct answer is option d) 17/36.
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In the MBA Programme of a B - School, there are two sections A and B. 1/4th of the students in Section A and 4/9th of the students in section B are girls. If two students are chosen at random, one each from section A and Section B as class representative, the probability that exactly one of the students chosen is a girl, is :a)23/72b)11/36c)5/12d)17/36Correct answer is option 'D'. Can you explain this answer?
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