a certain tennis player makes a successful first serve 71% of the time. assume that each serve is independent of the others. if she serves 5 times, what's the probability she gets c) at least 4 serves in? d) no more than 4 serves in?

Question

anonymous · Answer

tk help me with one after this?

tkhunny · Answer

Think about the probability that NO serve makes it in.

p(miss)*p(miss)*p(miss)*p(miss)*p(miss) = 0.29^5

What does that do for us?

anonymous · Answer

what? I dont understand

tkhunny · Answer

Ignoring the problem statement for a moment, what is the probability that ALL FIVE serves will be clinkers?

anonymous · Answer

.180

anonymous · Answer

I understand how to do the first two parts. I need help setting up the formula for c and d please

anonymous · Answer

Al teast 4 means  4 or 5 in this case.

tkhunny · Answer

No, that is the probability that all five will be good.
The probability that all 5 will be bad is 0.29^5 = 0.00205

Now, we have: 

p(0 good) = 0.00205
p(1 good) = 
p(2 good) = 
p(3 good) = 
p(4 good) = 
p(5 good) = 0.18042

That's just calculating the easy ones.  Does this enable us to answer the questions?

c) at least 4 serves in?  p(at least 4) = p(4) + p(5) -- Okay, so we don't quite know enough for this one.

d) no more than 4 serves in?  p(no more than 4) = p(0) + p(1) + p(2) + p(3) + p(4)
Aha!!!  We CAN solve this one.  p(no more than 4) = 1 - p(5) -- Do you see this?

anonymous · Answer

No more than 4 would mean 0, 1, 2, 3, or 4

anonymous · Answer

okay, got it 
thanks