There are four schools SSCI, SENG, SBM, and SHSS in UCLA. Suppose in UCLA the pr

Question

There are four schools SSCI, SENG, SBM, and SHSS in UCLA. Suppose in UCLA the probabilities that a randomly picked student is from SSCI, SENG, SBM, and SHSS are 1/6, 1/3, 1/3 and 1/6 . Suppose that the ratios of female and male students in SSCI, SENG, SBM, and SHSS are 1/2, 1/4, 3/1 and 1/1. Further assume that a student belongs to one of the four schools only (e.g., if a student belongs to SSCI, he/she does not belong to SENG, SBM, or SHSS).

(a) Suppose that 3 UCLA students are picked randomly. Find the probability that at least one of the 3 UCLA students are from SSCI or SENG.

(b) Suppose that n UCLA students are picked at random. Find the value of n such that the probability that at least one of the n UCLA students are from SSCI or SENG is greater than 0.9.

(c) Suppose a UCLA student is picked randomly. Given that the randomly picked student is a female, find the probability that this student is from SENG or SBM

Explanation / Answer

n UCLA students are picked at random.

probability that all the three of them are from SBM and SHSS is (1/3 + 1/6)n

So probability that at least one of them is from SSCI or SENG is 1-(1/3 + 1/6)n

a)

n = 3;

So probability that at least one of them is from SSCI or SENG is 1-(1/3 + 1/6)3   = 1-0.125 = 0.875

b)

given that probability that atleast one of the n students is from SSCI or SENG > 0.9

1-(1/3 + 1/6)n > 0.9

1/2n <0.1

2n > 10

n >= 4;

c)

let E be the event of a student being from SENG or SBM

E1 be the event of a student being from SENG, P(E1) = 1/3

E2 be the event of a student being from SBM, P(E2) = 1/3.

and F be the event of a student being female.

P(E|F) = [P(F|E1)P(E1) + P(F|E2)P(E2)]/P(F)

if girls to boys ratio in school is 1/n

number of girls would be k where k is some arbitrary number and

number of boys would be nk

total number of students = (n+1)k

probability that a random student picked from the school is girl = 1/(n+1)

P(F|E1) = 1/5

P(F|E2) = 3/4

P(F) = P(F|E1)P(E1)+P(F|E2)P(E2)+P(F|E3)P(E3)+P(F|E4)P(E4)

=[1/6*1/3 + 1/3*1/5 + 1/3*3/4 + 1/6*1/2] = 41/90

P(E|F) = [1/3*1/5 + 1/3*3/4] / (41/90) = 57/82 = 0.695

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