prob(b) = .51
prob(x) = .49 , where x is "not blue eyes"
one such case could be
xxbxxxx , which would be (.51)(.49^6)
but there are 7 such cases, so
prob(your event) = 7(.51)(.49^6)
= ....
in stats notation:
C(7,1)(.51)(.49)^6
The probability of a randomly drawn individual having blue eyes is 0.51. A) What is the probability that seven people drawn at random all have blue eyes?
(b) What is the probability that one of the sample of seven has blue eyes?
3 answers
I put the answer to part a) as 0.51^7 which gave me answer of 8.9741 to 4 dp.
For part B your working out gives an answer of 0.049 to 4 dp is this correct?
For part B your working out gives an answer of 0.049 to 4 dp is this correct?
I only answered part b) since a) is very simple
a) would be .51^7 = .0089741
I have no idea how yo got 8.9741, the digits are the same, but no probability can be greater than 1.
for b) the answer correct to 4 decimals is
.0494, your is correct only to 3 decimals
a) would be .51^7 = .0089741
I have no idea how yo got 8.9741, the digits are the same, but no probability can be greater than 1.
for b) the answer correct to 4 decimals is
.0494, your is correct only to 3 decimals