Question

Suppose that in IIT Madras, 40% of the students are girls. It is known that 70% of the boys play cricket, and 60% of the girls play cricket, while 40% of the boys play football, and 50% of the girls play football. Given that 33% of those who play football also play cricket. A student is randomly chosen, and it is known that the student plays cricket. What is the probability that the student plays football?

Answer 1

Applying Bayes' Theorem yields the probability that a cricket-playing student also plays football, approximately 0.23.

Let A be the event that a student plays cricket, and B be the event that a student plays football. We want to find P(B∣A), the probability that a student plays football given that they play cricket, using Bayes' Theorem. Given that 33% of those who play football also play cricket (P(A∣B)=0.33), and the proportions of cricket and football players among boys and girls, we calculate P(B) using the law of total probability:

P(B)=P(B∣Boy)⋅P(Boy)+P(B∣Girl)⋅P(Girl)

Substituting the given values, we find

P(B)=0.4⋅0.4+0.5⋅0.6=0.46.

Similarly,

P(A) is calculated using the law of total probability for cricket:

P(A)=P(A∣Boy)⋅P(Boy)+P(A∣Girl)⋅P(Girl)

Substituting values, we get

P(A)=0.7⋅0.6+0.6⋅0.4=0.66.

Now, applying Bayes' Theorem:

P(B∣A)= P(A∣B)⋅P(B)/P(A) =0.33⋅0.46/0.66 ≈0.23

So, the probability that a randomly chosen student who plays cricket also plays football is approximately 0.23.