consider two groups are students, where the first group has 7 women and 3 men an

Question

consider two groups are students, where the first group has 7 women and 3 men and the second group has 5 women and 5 men. Agroup is selected randomly and then students in the selected group are randomly arranged. Suppose the probability of choosing group 1 is p. (a) what is the probability that the first two are men (b) Given the first two are men, what is the probability that group 2 was selected.

Please expalin the answer. the answer was given by someone on this board as:

a) p*(3/10)* (2/9) + (1-p)*(5/10)*(4/9)
But I don't understand the terms (2/9), (4/9), 3/10, and 5/10

Please explain.

Explanation / Answer

 p*(3/10)* (2/9) + (1-p)*(5/10)*(4/9)

p*(3/10)* (2/9).....P1 = Probability for the first group

p is probability of choosing from group 1
probability that the first person is man = 3/10 .....3 men out of 10
afer selecting 1 man....9 persons left out of which 2 are men
=> probability that the second person is man = 2/9 .....2 men out of 9

 

 (1-p)*(5/10)*(4/9).......P2 = probability for the second group group

(1-p) is probability of choosing from group 2
probability that the first person is man = 5/10 .....5 men out of 10
afer selecting 1 man....9 persons left out of which 4 are men
=> probability that the second person is man = 4/9 .....4 men out of 9

=> Total Probability = P1 + P2 =  p*(3/10)* (2/9) + (1-p)*(5/10)*(4/9)

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.