Respuesta :
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)!)
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.