Binomial Probability problem

[jrpersia]

Here's a question I came across that I am looking for some help with. I think I am on the right track with parts a) & b). I am unsure whether or not I am missing something or even if I am proceeding correctly to solve part c. I hope you can help

The demand of a certain book in a bookstore on any given day is given by the following table:

y 0 1 2 3 4 5 6 7
P(y) 0.1 0.2 0.3 0.15 0.1 0.05 0.05 0.05

Find: a. The probability that on any given day the demand is more than 4

I think: P(y>4) = 1- P(y?4)
= 1 – [P(y=4) + P(y=3) + P(y=2) + P(y=1) + P(y=0)]
= 1 – [(0.1) + (0.15) + (0.3) + (0.2) + (0.1)]
= 1 - (0.85)
= (0.15)

Or

P(y>4) = P(y=5) + P(y=6) + P(y=7)
= (.05) + (.05) +(.05)
= (.15)

b. The probability that on any given day the demand is less than or equal to 2

Here I have: P(y?2) = P(y=0) + P(y=1) + P(y=2)
= (0.1) + (0.2) +(0.3)
= (.6)

c. On 5 randomly chosen days, what is the probability that 3 of the 5 days have demand more than 4

P(y) = 5!/(3!)(5-3)1 * (.15)3 * (.85)2
=.022

Does that sound about right?

 

[tkhunny]

Yes and no.

It sounds EXACTLY right until you produced the inexact, rounded final result. Try that again and don't round every intermedaite value to one significant digit. Did you used \(\displaystyle 0.15^{3}\;=\;0.003\)? Never do that again.

Why do you doubt? What deliberate errors did you make?