statistics question?

Question

if a problem reads

A professional baseball player believes that he is a good hitter, but his current batting average is only .250. That is he has only gotten hits in 25% of his at-bats. Find the probability that the player who really is a .250 hitter will get at least 18 hits in his next 50 at-bats

Now,
n=50
x=18
Probability of success or failure=.50

Do I include de .250 somewhere? or not?

Answer 1

a 0.250 hitter gets a hit 25% of the time so the success probability is .25 not .5.

Let X be the number of hits in 50 at bats
X ~ Binomial ( 50 , 0.25 )
P[X >= 18] = 1 - P[X <= 17]

Now either sum is going to take some time to compute.
but you get a solution of 0.055123

Another way to do this is to use the normal approximation to the binomial. You can use this approximation if np > 10 and n(1-p) > 10, in this case you have 12.5 and 37.5 so the approximation can be used.

X ~ Normal( 12.5, 9.375)
P[X >= 18] is approximated by P[X >=17.5 ] because of the continuity correction. You are after all using a continuous distribution to estimate probabilities of a discrete random variable.
P[X >=17.5 ] = P[ Z >= ( 17.5 - 12.5 ) / Sqrt( 9.375 )]
= P[ Z >= 1.63] = 0.0516

where Z is the standard normal

Answer 2

If I understand your problem correctly, a ".250 hitter" would have probability of success 25% and of failure 75% for every single hit. So the probability that the player who really is a .250 hitter will get EXACTLY18 hits in his next 50 at-bats is the probability of 32 failure and 18 success:
P(18) = (50, 18)(0.75)^32(0.25)^18
where (50, 18) is the number of different orders of success/failure hits in which 18 times are successful.
But the problem is asking for "at least" 18 hits. That means we need to including the probability of 19 hits, 20 hits, ...., 50 hits. Therefore:
P=P(18) + P(19) + P(20) + .... + P(50)
= (0.25)^18*{(50, 18)*(0.75)^32 + (50, 19)*(0.75)^31(0.25) + (50, 20)*(0.75)^30(0.25)^2 + (50, 21)*(0.75)^29(0.25)^3 + .... + (50, 50)*(0.25)^32}
= (0.25)^50*{(50, 18)*3^32 + (50, 19)*3^31 + (50, 20)*3^30 + (50, 21)*3^29 + .... + (50, 50)}
Another way to calculate is:
P = 1 - [P(1) + P(2) + P(3) + ... + P(17)]
I do not know how to get the closed form yet.