Question

In a dunking contest, T, J, and X are making shots one after the other, in that order. T makes (M) the basket 75% of the time, J misses the basket(M') 12% of the time, and X makes the basket 65% of the time

find the following probabilities

1. what is the probability that both T and J make the shot

2. that more make the shot than miss

3. that T and X miss their shots

4. what is the probability of X making his shot if we know that T and J miss their shot

5. what is the probability that at most 2 make the shot

Answer 1

Let,

The probability that T makes the shot = p1 = 0.75

The probability that J makes the shot = p2 = 0.88

The probability that X makes the shot = p3 = 0.65

Also, we can say that these events are independent of each other.

1. what is the probability that both T and J make the shot

P(T makes) * P(J makes) = 0.75*0.88 = 0.66

2. Probability that more make the shot than miss

Possible cases: 2 shots done 1 fails, and 3 shots done 0 fails

= P(T,J,X') + P(T,J',X) + P(T',J,X) + P(T,J,X) = 0.75*0.88*0.35 + 0.75*0.12*0.65 + 0.25*0.88*0.65 + 0.75*0.88*0.65

= 0.231 + 0.0585 + 0.143 + 0.429 = 0.8615

3. that T and X miss their shots

P(T',X') = 0.25*0.35 = 0.0875

4. what is the probability of X making his shot if we know that T and J miss their shot

Since the events are independent, hence P(X making shot ) is simply p3 = 0.65

5. what is the probability that at most 2 make the shot

P(atmost 2 makes the shot) = 1 - P(all three makes the shot) = 1 - 0.75*0.88*0.65 = 1 - 0.429 = 0.571