Respuesta :
Answer:
(A) 0.4419
(B) 0.2209
(C) 0.0265
(D) 0.2474
Step-by-step explanation:
Given,
Females = 7,
Males = 5,
Total candidates = 7 + 5 = 12,
The total ways of choosing any 5 candidates = [tex]^{12}C_5[/tex]
(A) Number of ways of choosing 3 females and 2 males = [tex]^7C_3\times ^5C_2[/tex]
Since,
[tex]\text{Probability}=\frac{\text{Favourable outcomes}}{\text{Total outcomes}}[/tex]
Thus, the probability of choosing 3 females and 2 males = [tex]\frac{ ^7C_3\times ^5C_2}{^{12}C_5}[/tex]
[tex]=\frac{\frac{7!}{3!4!}\times \frac{5!}{2!3!}}{\frac{12!}{5!7!}}[/tex]
[tex]=\frac{350}{792}[/tex]
[tex]=0.4419[/tex]
Similarly,
(B)
The probability of choosing 4 females and 1 male = [tex]\frac{ ^7C_4\times ^5C_1}{^{12}C_5}[/tex]
[tex]=\frac{\frac{7!}{4!3!}\times 5}{\frac{12!}{5!7!}}[/tex]
[tex]=\farc{35}{792}[/tex]
= 0.2209
(C)
The probability of choosing 5 females = [tex]\frac{ ^7C_5\times ^5C_0}{^{12}C_5}[/tex]
[tex]=\frac{\frac{7!}{5!2!}}{\frac{12!}{5!7!}}[/tex]
[tex]=\farc{21}{792}[/tex]
= 0.0265
(D)
The probability of choosing at least 4 females = [tex]\frac{ ^7C_4\times ^5C_1+^7C_5\times ^5C_0}{^{12}C_5}[/tex]
[tex]=\frac{\frac{7!}{4!3!}\times 5+\frac{7!}{5!2!}\times 1}{\frac{12!}{5!}{7!}}[/tex]
[tex]=\frac{196}{792}[/tex]
[tex]=0.2474[/tex]
