Question

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 10 participants there have been only 2 women. if participants are picked randomly, what is the probability of getting 2 or fewer women when 10 people are picked

Answer 1

Answer:

The probability of getting 2 or fewer women when 10 people are picked

P(x≤ 2) = $\frac{56}{1024} = 0.0546$

Step-by-step explanation:

The participants in a television quiz show are picked from a large pool of applicants with approximately equal numbers of men and women

That is probability of participants of television quiz of equal numbers of men and women that is 50 %of men and 50% of women.

p = 50/100 = 1/2

q = 1-p = 1 - 1/2 = 1/2

n =10

we will use binomial distribution $p(x =r) = n_{cr} p^{r} q^{n-r}$

The probability of getting 2 or fewer women when 10 people are picked

P(x≤ 2) = P(x=0)+P(x=1)+P(x=2)

          = $10_{c0} \frac{1}{2} ^{0} (\frac{1}{2}) ^{10-0}+ 10_{c1} \frac{1}{2} ^{1} (\frac{1}{2}) ^{10-1$   + $10_{c2} \frac{1}{2} ^{2} (\frac{1}{2}) ^{10-2$

          by using formula $n_{cr} =\frac{n!}{(n-r)!r!}$

on simplification we get

     = $10_{c0} (\frac{1}{2}) ^{10}+ 10_{c1} (\frac{1}{2}) ^{10)$ + $10_{c2} (\frac{1}{2}) ^{10)$

    = $1 (\frac{1}{2}) ^{10}+ 10(\frac{1}{2}) ^{10)$+$45 (\frac{1}{2}) ^{10}$

  = $\frac{56}{1024} = 0.0546$

Conclusion:-

The probability of getting 2 or fewer women when 10 people are picked

P(x≤ 2) = $\frac{56}{1024} = 0.0546$