Question

[JSOM Home Run Derby] A home run derby in baseball consists of a batter taking the stage, and facing a series of pitches (balls thrown towards the batter), attempting to hit a series of pitches (balls thrown towards the batter) them for home runs. The Jindal School of Management is organizing its own home run derby for faculty members to raise fund for the future classroom. In the JSOM version, each batter gets exactly 15 attempts. The first batter is Prof. X, who, having excelled at both baseball and cricket in childhood, is a particularly good batter. It is estimated that on any given attempt, Prof. X has a 60% chance of hitting a ball for a home run. The outcome of each attempt is independent of all others. Let H denote the number of successful home runs hit by Prof. X over the course of the 15 attempts. For every ball Prof. H hits for a home run, Nick Jackman (a wealthy former alum) will donate $1,000 to the JSOM’s local charity. Also, if Prof. X hits 10 or more home runs, Jackman will donate an additional $10,000.

What is the probability that Jackman’s total donation is greater than $7,500? (a) 0.213 (b) 0.390 (c) 0.610 (d) 0.787

Answer 1

Probability of hitting home run (p) = 0.6

Probability of not hitting home run (q) = 1-0.6 =0.4

We can see that this question can be done using binomial distribution as there are only two possible outcomes.

And now we know that $1000 is given for every home run. To have an amount greater than $7500, we Professor X needs to hit atleast 8 home runs.

Let X be the number of times home run is hit.

And when X follows binomial distribution, P(X=x) = nCx*p^x*q^(n-x), where n is total observation, p is probability of an event occurring and q = 1-p

So the probability of having donation of atleast $7500 = P(X=8) + P(X=9)+P(X=10)+P(X=11)+P(X=12)+P(X=13) + P(X=14) + P( X=15)

= 15C8*0.6⁸*0.4⁷+15C9*0.6⁹*0.4⁶+15C10*0.6¹⁰*0.4⁵+15C11*0.6¹¹*0.4⁴​​​​​​​+15C12*0.6¹²*0.4³​​​​+15C13*0.6¹³*0.4²​​​​​​​+ 15C14*0.6¹⁴*0.4¹ + 15C15*0.6¹⁵*0.4⁰

= 0.787

So option d is the correct answer.