Answer:
a) [tex] P(F\cap F) = P(F)*P(F) = 0.55*0.55= 0.3025[/tex]
b) For this case we want the probability that at least one of the two students is a woman.
At least mean that can be 1 one women or 2. Let X the number of women selected:
[tex] PX\geq 1) = P(X=1) +P(X=2)[/tex]
[tex] P(X=1)= 0.55*0.45 +0.45*0.55= 0.495[/tex]
[tex] P(X=2) =0.55*0.55= 0.3025[/tex]
So then the final probability would be:
[tex] PX\geq 1) = 0.495 +0.3025 = 0.7975[/tex]
Step-by-step explanation:
For this case we have the situation described in the figure attached.
We define the following notation:
F= Select a random student women
M= Select a random student male
We have defined the prior probabilities:
[tex] P(F)= 0.55, P(M) =1-0.55= 0.45[/tex]
We can define the sample space like this:
[tex] S = [MM, MF, FM, FF][/tex]
For this case we assume that the selection on each trial are independnet events.
Part a
We want to find this probability [tex] P(FF)[/tex], using the indpendence condition we can do this:
[tex] P(F\cap F) = P(F)*P(F) = 0.55*0.55= 0.3025[/tex]
Part b
For this case we want the probability that at least one of the two students is a woman.
At least mean that can be 1 one women or 2. Let X the number of women selected:
[tex] PX\geq 1) = P(X=1) +P(X=2)[/tex]
[tex] P(X=1)= 0.55*0.45 +0.45*0.55= 0.495[/tex]
[tex] P(X=2) =0.55*0.55= 0.3025[/tex]
So then the final probability would be:
[tex] PX\geq 1) = 0.495 +0.3025 = 0.7975[/tex]
