Question

In a study conducted by the Bank of America, it was found that 34% of young Millenials say that they sleep with their smartphone on their bed. If four Millenials are selected at random what is the probability that:
a) None of them sleep with their phone on their bed?
b) Exactly two sleep with their phone on their bed?
c) At least two of them sleep with their phone on the bed?

What I know: n=4, p=34/100 q= 66/100 from 1-34/100

for a I got .1897
p(x=0) c(4,0) (34.100)^0 (66/100)^4

and for part b I got .3021
p(x=2) c(4,2) (34/100)^2 (66/100)^2


but I'm stuck on part c.

Answer 1

p yes = .34
q no=1-p = 1-.34 = .66

a. yes n = 4,
none of them C(4,0) .34^0 * .66^4 = 1 * 1 * .66^4 = .1897 yes

b. C(4,2) .34^2 .66^2 = 6 * .1156* .4356 = .3021 yes

c. prob( 2 or 3 or 4) = 1 - (just one + none )
prob of zero = .1897 we know from part a
prob of one = C(4,1).34^1 * .66^3 = 4 * .34* .2875 = .3910
sum = .5807
1 - .5807 = .4193

Answer 2

You sure have raced through this subject in two days !