Question

In: Statistics and Probability

1. The number of raisins in buns has Poisson distribution with an average of 5 raisins...

1. The number of raisins in buns has Poisson distribution with an average of 5 raisins per bun.

a) Bob buys a bun every day. What is the expected number of buns he buys before he finds a bun with no raisins at all?

b) Joe bought a bun for breakfast and cut it into two equal parts. What is the probability that at least one of these parts has no raisins in it?

c) Emily bought seven buns in the last month. What is the probability that fewer than two of these buns had at least three raisins in them?

Solutions

Expert Solution

Question 1

Here the number of raisins in buns has a poisson distribution with parameter = 5

(a) Here we have to find first the probability that he finds a buns with no raising at all

P(x = 0) = e-5 = 0.0067

so here expected number of buns he has to buy to finds a bun with no raisins at all = 1/0.0067= 148.41 or 148 buns

(b) Here he cut the bun into two equal parts.

That means the expected number of bun for breakfast = = 2.5

so here we have to find

P(at least one of these parts has no raisins in it) = 1 - P(none of the part has no raisin in it)

P(none of the part has no raising in it) = P(First half have raisin) * P(Second half have raisin)

P(First half have raisin) = 1 - P(First hald doesn't have raisin) = 1 - e-2.5 = 0.9179

P(none of the part has no raising in it) = 0.9179 * 0.9179 = 0.8426

P(at least one of these parts has no raisins in it) = 1 - 0.8426 = 0.1574

(c) Here

Probability that a bud had at least three raisins = P(x 3) = 1 - P(x 2)

using POISSON distribution calculator

= 1 - POISSON (x 2 ; = 5)

= 1 - 0.1247 = 0.8753

so here now n = 7 and p = 0.8753

if y is the number of buns out of 7 that have at least three raisins in them then

y ~ BINOMIAL (n = 7, p = 0.8753)

we have to find

P(x < 2) = BINOMIAL (x < 2 ; n = 7; p = 0.8753)

using BINOMIAL distribution calculator

P(x < 2) = BINOMIAL (x < 2 ; n = 7; p = 0.8753) = 0.0000234


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