SOLUTION: 4.24. There are two urns containing pens and pencils. Urn #1 contains three pens and seven pencils and Urn #2 contains eight pens and four pencils. (a) An urn is chosen at random

Algebra ->  Probability-and-statistics -> SOLUTION: 4.24. There are two urns containing pens and pencils. Urn #1 contains three pens and seven pencils and Urn #2 contains eight pens and four pencils. (a) An urn is chosen at random      Log On


   

Question 1209842: 4.24. There are two urns containing pens and pencils. Urn #1 contains three
pens and seven pencils and Urn #2 contains eight pens and four pencils.
(a) An urn is chosen at random and an object is drawn. What is the proba-
bility that it is a pencil?
(b) An urn is chosen at random and an object is drawn. If the object drawn
is a pencil, what is the probability that it came from Urn #1?
(c) If an urn is chosen at random and two objects are drawn simultaneously,
what is the probability that both are pencils?
Found 2 solutions by ikleyn, ggvindy:
Answer by ikleyn(52855) About Me  (Show Source):
You can put this solution on YOUR website!
.
4.24. There are two urns containing pens and pencils. Urn #1 contains three
pens and seven pencils and Urn #2 contains eight pens and four pencils.
(a) An urn is chosen at random and an object is drawn. What is the proba-
bility that it is a pencil?
(b) An urn is chosen at random and an object is drawn. If the object drawn
is a pencil, what is the probability that it came from Urn #1?
(c) If an urn is chosen at random and two objects are drawn simultaneously from this urn,
what is the probability that both are pencils?
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        I edited your post to make the problem as clear as it should be.
                            My editing is underlined. 


(a)  P(a) = %281%2F2%29%2A%287%2F10%29 + %281%2F2%29%2A%284%2F12%29 = 7%2F20 + 4%2F24 = 42%2F120 + 20%2F120 = 62%2F120 = 31%2F60.



(b)  P(b) = 7%2F%287%2B4%29 = 7%2F11.



(c)  P(c) = %281%2F2%29%2A%287%2F10%29%2A%286%2F9%29 + %281%2F2%29%2A%284%2F12%29%2A%283%2F11%29 = %287%2F10%29%2A%281%2F3%29 + %281%2F6%29%2A%283%2F11%29 = 7%2F30 + 3%2F66 = 7%2F30 + 1%2F22 = %287%2A22%2B30%29%2F%2830%2A22%29 = 184%2F%2830%2A22%29 = 46%2F165.

Solved.



Answer by ggvindy(1) About Me  (Show Source):
You can put this solution on YOUR website!
- Let U1 be the event that Urn 1 is chosen
- Let U2 be the event that Urn 2 is chosen
- Let Pencil be the event that a Pencil is chosen
- Let Pen be the event that a Pen is chosen
- P(U1) = P(U2) = 1/2
- P(Pencil|U1) = 7/10
- P(Pencil|U2) = 4/12
==============================================
(a)
- P(U1) * P(Pencil|U1) + P(U2) * P(Pencil|U2)
=1/2 . 7/10 + 1/2 . 4/12 = 31/60 Ans
==============================================
(b)
- Here we need to apply Bayes theorem where we need to calculate Probability of Urn 1 is chose given that the Pencil is chosen : P(U1|Pencil)

- As per Bayes Theorem, P(U1|Pencil) = [P(Pencil|U1) * P(U1)] / Total Probability
i.e. P(U1|Pencil) = [P(Pencil|U1) * P(U1)] / [P(Pencil|U1) * P(U1)] +[P(Pencil|U2) * P(U2)]
i.e. P(U1|Pencil) = [7/10 . 1/2] / [7/10 . 1/2 + 4/12 . 1/2] = 21/31 Ans
==============================
(c)
- Probability of choosing Urn 1 P(U1) = 1/2
- Probability of choosing 2 Pencils from Urn 1 = P(Pencil2U1) = 7C2/10C2 where C=possible Combinations
- Probability of choosing Urn 2 P(U2) = 1/2
- Probability of choosing 2 Pencils from Urn 2 = P(Pencil2U2) = 4C2/12C2 where C=possible Combinations
Therefore answer =
1/2 . 7C2/10C2 + 1/2 . 4C2/12C2 = 1/2 . 7/15 + 1/2 . 1/11 = 46/165 Ans