Given the probability that a certain tennis player will serve an ace is 1/4 and serves 5 times.

a) write a probability distribution of the number of times the player serves an ace.
b) what is the probability of serving less than 3 aces?
c) what is the probability of serving greater than 1 ace?

a) would it look something like this?
x 1 2 3 4 5
P(x=x) 1/4 1/4 1/4 1/4 1/4

possible outcomes in the 5 serves:

no aces -- C(5,0) (1/4)^0 (3/4)^5 = 243/1024
1 ace -- C(5,1) (1/4)(3/4)^4 = 405/1024
2 aces -- C(5,2)(1/4)^2 (3/4)^3 = 270/1024
3 aces -- C(5,3)(1/4)^3 (3/4)^2 = 90/1024
4 aces -- C(5,4)(1/4)^4 (3/4) = 15/1024
5 aces = C(5,5)(1/4)^5 (3/4)^0 = 1/1024
(notice they add up to 1 )

for b) and c) , just add up the appropriate results

e.g. for the probability of serving less than 3 aces:
add up no aces, 1 ace and 2 aces

Yes, you're on the right track for part (a). The probability distribution of the number of times the player serves an ace would look something like this:

x 0 1 2 3 4 5
P(x=x) 3/4 1/4 1/4 1/4 1/4 1/4

In this case, x represents the number of aces served and P(x=x) represents the probability of serving x aces.

For part (b), you are asked to find the probability of serving less than 3 aces. To solve this, you sum up the probabilities of serving 0 or 1 ace:

P(x<3) = P(x=0) + P(x=1) = 3/4 + 1/4 = 4/4 = 1

So, the probability of serving less than 3 aces is 1.

For part (c), you need to calculate the probability of serving greater than 1 ace. To solve this, you sum up the probabilities of serving 2, 3, 4, or 5 aces:

P(x>1) = P(x=2) + P(x=3) + P(x=4) + P(x=5) = 1/4 + 1/4 + 1/4 + 1/4 = 4/4 = 1

So, the probability of serving greater than 1 ace is also 1.