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Len [333]
3 years ago
6

Suppose thatAnn selects a ball by first picking one of two boxes at random and then selecting a ball from this box. The first bo

x contains three orange balls and four black balls, and the second box contains five orange balls and six black balls. What is the probability that Ann picked a ball from the second box if she has selected an orange ball?
Mathematics
1 answer:
Natalka [10]3 years ago
3 0

Answer:

35/68

Step-by-step explanation:

The first box contains 3 orange balls and 4 black balls.

The second box contains 5 orange balls and 6 black balls

Total number of boxes = 2

Let Pr(F) be the probability that the first box is picked.

Let Pr(S) be the probability that the second ball is picked

Let O be the orange ball selected

Pr(F) = 1/2

Pr(S) = 1/2

Pr(O|S) = n(O) / n(S)

= 5/ 6+5

5/11

Pr(O|F) = n(O) / n(F)

= 3/ 4+3

= 3/7

Pr(S|O) = [Pr( O|S).P(S)] / [Pr( O|S).P(S) + Pr( O|F).P(F)]

= (5/11*1/2) / (5/11*1/2) +(3/7*1/2(

= (5/22) / (5/22 + 3/14)

= (5/22) / (136/308)

= 35/68

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A librarian has 4 identical copies of Hamlet, 3 identical copies of Macbeth, 2 identical copies of Romeo and Juliet, and one cop
lesantik [10]

Answer:

The number of distinct arrangements is <em>12600</em><em>.</em>

Step-by-step explanation:

This is a permutation type of question and therefore the number of distinguishable permutations is:

n!/(n₁! n₂! n₃! ... nₓ!)

where

  • n₁, n₂, n₃ ... is the number of arrangements for each object
  • n is the number of objects
  • nₓ is the number of arrangements for the last object

In this case

  • n₁ is the identical copies of Hamlet
  • n₂ is the identical copies of Macbeth
  • n₃ is the identical copies of Romeo and Juliet
  • nₓ = n₄ is the one copy of Midsummer's Night Dream

Therefore,

<em>Number of distinct arrangements =  10!/(4! × 3! × 2! × 1!)</em>

<em>                                                         = </em><em>12600 ways</em>

<em />

Thus, the number of distinct arrangements is <em>12600</em><em>.</em>

4 0
3 years ago
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