Question

Bob is a high school basketball player. he is a 70% free throw shooter. that means his probability of making a free throw is 0.70. during the season, what is the probability that bob makes his third free throw on his fifth shot? what is e(x)?

Answer 1

Since Bob makes 70% of free throws, P (probability of success)=0.70. x (number of trials) is five because there are a total of five shots. R (number of successes)=3 because we are determining whether or not the third free throw will be successful. We will use the following formula b*(x; r, P) = x-1Cr-1 * Pr * Qx - r b*(5; 3, 0.7) = 4C2 * 0.73 * 0.32 b*(5; 3, 0.7) = 6 * 0.343 * 0.09 = 0.18522 The probability he will make the shot is 0.18522. For the purpose of this problem, X=0.70

Answer 2

P =Probability of making free throw by Bob = 70 % = 0.70

Q=Probability that bob will be not able to make a free throw = 1 - 0.70= 0.30

Probability, that bob makes his third free throw on his fifth shot

                  = Taking fifth throw as free throw and Remaining 4 throws in which (two are free throw=P) and (2 are not free throw=Q)

          = $\frac{4!}{2!\times2!}=\frac{4\times3\times2}{4}=6$ ways

        = 6×Q²× P³

        = 6 × (0.30)²×(0.70)³

   e(x)  =  0.18522 (Approx)