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A charity is conducting a raffle, and it sells one ticket each to 35 women and 21 men. If 5 winners are randomly selected, what is the probability that they are all men? Round to five decimal places. 0.07776 0.09537 0.00533 0.06268

Respuesta :

sqdancefan
A total of 35+21 = 56 tickets were sold. The probability of 5 men winning is the ratio
  C(21, 5)/C(56, 5) = 20,349/3,819,816 ≈ 0.00533

_____
C(n, k) is the number of ways k items can be chosen from n items. Its value is
  = n!/(k!*(n-k)!)

Answer:

0.00533

Step-by-step explanation:

Given : A charity is conducting a raffle, and it sells one ticket each to 35 women and 21 men.

To Find:If 5 winners are randomly selected, what is the probability that they are all men?

Solution:

Number of female contestants = 35

Number of male contestants = 21

Total number of contestants = 35+21=56

Now we are given that 5 winners are randomly selected

So, Probability that they are all men=[tex]\frac{^{21}C_5}{^{56}C_5}[/tex]

                                                           =[tex]\frac{\frac{21!}{5!\left(21-5\right)!}}{\frac{56!}{5!\left(56-5\right)!}}[/tex]

                                                           =[tex]0.00533[/tex]

Thus the probability that they are all men if 5 winners are randomly selected is 0.00533.

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