the participants in a television quiz show are picked from a large pool of applicants with approximately equal numbers of men and women. Among the last 11 participants there have been only 2 women. if participants are picked randomly, what is theprobability that out of 11 people we get:

a) 2 or fewer women

b) At least 2 woman

C) nomore than one woman

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest "number of women", on this case we now that:

[tex]X \sim Binom(n=11, p=0.5)[/tex]

The probability mass function for the Binomial distribution is given as:

[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]

Where (nCx) means combinatory and it's given by this formula:

[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]

Part a

For this case we want to find this probability:

[tex] P(X \leq 2)= P(X=0)+P(X=1)+P(X=2)[/tex]

[tex]P(X=0)=(11C0)(0.5)^0 (1-0.5)^{11-0}=0.00049[/tex]

[tex]P(X=1)=(11C0)(0.5)^1 (1-0.5)^{11-1}=0.0054[/tex]

[tex]P(X=2)=(11C0)(0.5)^2 (1-0.5)^{11-2}=0.027[/tex]

And adding we got:

[tex] P(X \leq 2)= 0.033[/tex]

Part b

For this case we want this probability:

[tex] P(X \geq 2)[/tex]

And we can use the complement rule and we got:

[tex] P(X \geq 2)= 1-P(X<2) = 1-P(X\leq 1)=1-[P(X=0) +P(X=1)][/tex]

And replacing we got:

[tex] P(X \geq 2)= =1-[0.00049 +0.0054] = 0.994[/tex]

Part c

For this case we want this probability:

[tex] P(X \leq 1)[/tex]

And we can use the complement rule and we got:

[tex] P(X \leq 1)= 1-P(X<2) = P(X=0) +P(X=1)[/tex]

And replacing we got:

[tex] P(X \leq 1)=0.00049 +0.0054= 0.0059[/tex]