- #1

diracdelta

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## Homework Statement

We have 6 new and 4 used products in a box.

We draw randomly 2 products and used them for a while. then return them back into the box.

After that, again we draw 2 products.

a) What is probability that at least one is new?

b) What is probability that at least one is new if we found out that in first draw two used were not drawn out.

## The Attempt at a Solution

a)

A={ we drawn out 2 used products in second draw}

H

_{1}={ we drawn out 2 new products in first draw}

H

_{2}={ we drawn out 1 new and 1 used product in first draw}

H

_{3}={ we drawn out 1 used and 1 new product in first draw}

H

_{4}={ we drawn out 2 used product in first draw}

P(H

_{1})=6/10 * 5/9 = 1/3

P(H

_{2})=6/10 * 4/9 = 24/90

P(H

_{3})=4/10 * 6/9 = 24/90

P(H

_{4})=4/10 * 3/9 = 12/90

P(A|H

_{1})=6/10 * 5/9 = 1/3

P(A|H

_{2})=5/10 * 4/9 = 2/9

P(A|H

_{3})=4/10 * 5/9 = 2/9

P(A|H

_{4})=4/10 * 3/9= 12/90

P(A) = 1/3 * 1/3 + 24/90 * 2/9 + 24/90 * 2/9 + 12/90 * 12/90

P(A) = 0.24

P(at least one new) = 1- P(A) = 0.76

b)

A={ we drawn out 2 used products in second draw}

H

_{1}={ we drawn out 2 new products in first draw}

H

_{2}={ we drawn out 1 new and 1 used product in first draw}

H

_{3}={ we drawn out 1 used and 1 new product in first draw}

P(H

_{1})=6/10 * 5/9 = 1/3

P(H

_{2})=6/10 * 4/9 = 24/90

P(H

_{3})=4/10 * 6/9 = 24/90

P(A|H

_{1})=6/10 * 5/9 = 1/3

P(A|H

_{2})=5/10 * 4/9 = 2/9

P(A|H

_{3})=4/10 * 5/9 = 2/9

P(A) = 1/9 + 2/9 * 24/90 + 2/9 * 24/90

P(A) = 0.2296

P(at least one new) = 1- P(A) = 0.7703Is this correct solution?

My apologies if bad english, and if unpractical in absence of latex.

Thanks!