Question

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1. A host gives a scratchie to each of the 100 guests at his party. Each scratchie has probability p of winning some prize independent of the others) (a) What is the probability that exactly 10 guests win a prize? (b) What is the expected number of guests who win a prize? (c) What is the probability that at least 3 guests win priz Make your final answer a reasonably short expression that does not use sigma notation Answer only required for (a) and (b), short explanation required for (c)] [6)

Answer 1

(a)

Binomial Distribution:

n = Number of trials = 100

p = Probability of success in a single trial = p

So,

q = 1 - p

Binomial Distribution: n = Number of trials = 100 p = Probability of success in a single trial = p So, q = 1 - p P(X = 10) = (100 0) times (1 - p)^90 times p^10 = 1.7310 times 10^13 times (1 - p)^90 times p^10 so, 1.7310 times 10^13 times (1 - p)^90 times p^10 Expected Number = n p = 100 p So, 100 p p(x greaterthanorequalto 3) = 1 - [p(x = 0) + p(x = 1) + p(x = 2)] p(X = 0) = (100 0) times (1 - p)^100 times p^0 = 1 times (1 - p)^100 times 1 = (1 - p)^100 P(X = 1) = (100 1) times (1 - p)^99 times p^1 = 100 times (1 - p)^99 times p = 100(1 - p)^99 p P(X = 2) = (100 2) times (1 - p)^98 times p^2 = 4950 times (1 - p)^98 times p^2 = 4950(1 - p)^98 p^2 so P(X greaterthanorequalto 3) = 1 - [1

So,

Answer is:

(b)

Expected Number = n p = 100 p

So,

Answer is:

100 p

(c)

P(X $\geq$ 3) = 1 - [P(X =0) + P(X = 1) + P(X = 2)]

So