On October 6, 2010, Roy Halladay of the Philadelphia Phillies pitched the second
ID: 3337146 • Letter: O
Question
On October 6, 2010, Roy Halladay of the Philadelphia Phillies pitched the second post-season no-hitter in the history of Major League Baseball.
a) Phillips got three hits in four turns at bat. Remembering that his batting average (his change of getting a hit) this year was 0.275, and assuming again that the probability of Phillips getting a hit is constant at that level and if each chance is independent, what is the probability that Phillips gets less than four hits in eight turns at bat?
b) Another part of that process was striking out the Cincinnati Reds' third baseman, Scott Rolen, three times. This year, Rolen struck out in 82 of his 537 plate appearances. If the probability of Rolen striking out is constant at that level and if each chance is independent, what is the probability that Rolen strikes out three times in a row?
c) During that game, Halladay came up to bat three times and got one hit. This year, Halladay's batting average was 0.141. If the probability of Halladay getting a hit is constant at that level and if each chance is independent, what is the probability that Halladay would get exactly one hit in three tries?
Explanation / Answer
Solution- (a) Let us denotethe number of hits by phillips by X which follows binomial distribution with parameters
number of trials ( n) = 8 and probability of success (p) = .275
Required probability = P(X<4)
= P(X=0) + P(X=1) + P(X=2)+ P(X=3)
= .8487
(b) Probability of hitting a target is 82/537, then probability of hiting a target 3 times in a row is ( 82/537 )^3
=.00356
(c) Let us denotethe number of hits by halladay by X which follows binomial distribution with parameters
number of trials ( n) = 3 and probability of success (p) = .141
Required probability = P(X=1)
= 3c1 (.141)^1 (1-.141)^2
=.312
ANSWERS
TY!
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