Probability Problems: Find the Answer ASAP

[rakesh1988]

1. box 'a' contains 2 white and 4 black balls, box 'b' contains 5 white and 7 black balls. a ball is transferred from a to b. then a ball is drawn from b. I) what is the probability that it is white? II) and it is black?

2.find the prob that in a random arrangement of the word UNIVERSITY, 2I's do not come together

pls help me in solving these problem ASAP... its urgent

 

[CompuChip]

Welcome to PF rakesh. Unfortunately you have overwritten that terribly handy template that comes up when you try to post here, so I do not know what you have tried already. But let me give you a hint:
P(drawing a white ball) = P(drawing a white ball | a white ball was transferred from a to b) + P(drawing a white ball | a black ball was transferred from a to b).

For the second one, I suggest checking how many rearrangements of the letters there are and how many there are in which the two I's are together.

 

[rakesh1988]

Solution:

box A contains
2W 4B

box B contains
5W 7B

Step 1)

When a ball is transferred from A to B

= For box A: P(w) = Number of chances you might get white (Events)/ Total number chances = 2 / 6 = 1/3

= For box A: P(b) = 4 / 6 = 2 / 3

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Step 2)

Now, assuming White is Transferred from A to B
B contains:
6W 7B

So For box B: p(w)= 6 / 13

Now, assuming Black is Transferred from A to B
B contains:
5W 8B

So For box B: p(b)= 8 / 13
-----------------------------------------------------------------------

Step 3:

Probability of box B is dependent on probability of box A.

So formula for dependent or probability is

Total Probability = P(A).P(B|A)

P(B|A) means Probability of box B Assuming Probability of box A

So

Ans
1) Probablity that is white = Box A P(w) * Box A P(w) = 1/3 * 6/13 = 6/39 = 2/13

2) Probablity that is black = Box A P(b) * Box A P(b) = 2/3 * 8/13 = 16/39


is this ok?? pls help

 

[CompuChip]

Step 1 and 2 look correct, although your answer is not going entirely as it should

Let's look at the probability P(W) that a ball you draw is white. Then
P(W) = P(W | W) + P(W | B)
where P(W | A) means: the probability of drawing a white, given that color A (either W(hite) or B(lack)) was transferred. In step 1 and 2 you have reasoned that, for example
P(W | W) = P(transferring a white ball from A to B) * P(drawing a white ball from B, after a white ball was added) = 2/6 * 6/13
and
P(W | B) = P(transferring a black ball from A to B) * P(drawing a white ball from B, after a black ball was added) = 4/6 * 5/13.

So you get
P(W) = 2/6 * 6/13 + 4/6 * 5/13 = (6 + 10) / 39 = 16/39
which seems to be what you have with P(B).

Also note how your answer cannot be correct: whatever ball you transfer, when you draw from B it must be either black or white. So P(W) + P(B) = 1, which your answer does not satisfy (2/13 + 16/39 = 22/39 =/= 1).

I think your mistake is in the formula
P(B) = P(B | A) . P(A)
which should actually be
P(B) = Sum[over all possibilities in A]( P(B | A) P(A) )
so in this case
P(B) = P(2W,5B in A) P(draw white from B | 2W, 5B in A) + P(3W, 4B in A) P(draw white from B | 3W, 4B in A).