Mathematics, 16.07.2020 06:01 murtaghliam1

A math class consists of 25 students, 14 female and 11 male. Three students are selected at random to participate in a probability experiment. Compute the probability that a. a male is selected, then two females. b. a female is selected, then two males. c. two females are selected, then one male. d. three males are selected. e. three females are selected.

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Answer from: anglacx5465

(a) The probability that a male is selected, then two females is 0.4352.

(b) The probability that a female is selected, then two males is 0.3348.

(d) The probability that three males are selected is 0.0717.

(e) The probability that three females are selected is 0.1583.

Step-by-step explanation:

We are given that a math class consists of 25 students, 14 female and 11 male. Three students are selected at random to participate in a probability experiment.

(a) The probability that a male is selected, then two females is given by;

Number of ways of selecting a male from a total of 11 male = $^{11}C_1$

Number of ways of selecting two female from a total of 14 female = $^{14}C_2$

Total number of ways of selecting 3 students from a total of 25 = $^{25}C_3$

So, the required probability = $\frac{^{11}C_1 \times ^{14}C_2}{^{25}C_3}$

= $\frac{\frac{11!}{1! \times 10!} \times \frac{14!}{2! \times 12!} }{\frac{25!}{3! \times 22!} }$ { $\because ^{n}C_r = \frac{n!}{r! \times (n-r)!}$ }

= $\frac{1001}{2300}$ = 0.4352

(b) The probability that a female is selected, then two males is given by;

Number of ways of selecting a female from a total of 14 female = $^{14}C_1$

Number of ways of selecting two males from a total of 11 male = $^{11}C_2$

Total number of ways of selecting 3 students from a total of 25 = $^{25}C_3$

So, the required probability = $\frac{^{14}C_1 \times ^{11}C_2}{^{25}C_3}$

= $\frac{\frac{14!}{1! \times 13!} \times \frac{11!}{2! \times 9!} }{\frac{25!}{3! \times 22!} }$ { $\because ^{n}C_r = \frac{n!}{r! \times (n-r)!}$ }

= $\frac{770}{2300}$ = 0.3348

Number of ways of selecting two females from a total of 14 female = $^{14}C_2$

Number of ways of selecting one male from a total of 11 male = $^{11}C_1$

Total number of ways of selecting 3 students from a total of 25 = $^{25}C_3$

So, the required probability = $\frac{^{14}C_2 \times ^{11}C_1}{^{25}C_3}$

= $\frac{\frac{14!}{2! \times 12!} \times \frac{11!}{1! \times 10!} }{\frac{25!}{3! \times 22!} }$ { $\because ^{n}C_r = \frac{n!}{r! \times (n-r)!}$ }

= $\frac{1001}{2300}$ = 0.4352

(d) The probability that three males are selected is given by;

Number of ways of selecting three males from a total of 11 male = $^{11}C_3$

Total number of ways of selecting 3 students from a total of 25 = $^{25}C_3$

So, the required probability = $\frac{^{11}C_3}{^{25}C_3}$

= $\frac{ \frac{11!}{3! \times 8!} }{\frac{25!}{3! \times 22!} }$ { $\because ^{n}C_r = \frac{n!}{r! \times (n-r)!}$ }

= $\frac{165}{2300}$ = 0.0717

(e) The probability that three females are selected is given by;

Number of ways of selecting three females from a total of 14 female = $^{14}C_3$

Total number of ways of selecting 3 students from a total of 25 = $^{25}C_3$

So, the required probability = $\frac{^{14}C_3}{^{25}C_3}$

= $\frac{ \frac{14!}{3! \times 11!} }{\frac{25!}{3! \times 22!} }$ { $\because ^{n}C_r = \frac{n!}{r! \times (n-r)!}$ }

= $\frac{364}{2300}$ = 0.1583

Answer from: anaalashay

(a) The probability that a male is selected, then two females is 0.4352.

(b) The probability that a female is selected, then two males is 0.3348.

(d) The probability that three males are selected is 0.0717.

(e) The probability that three females are selected is 0.1583.

Answer from: Quest

250

step-by-step explanation:

Answer from: Quest

d i think

step-by-step explanation:

Me plz i just need this one last test to pass

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