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# Statistics homework need help please

Sun 12 Mar, 2017 04:52 pm
Bag A has 6 black balls and 1 white ball. Bag B has 14 black balls and 1 white ball. Jane has Bag A and Dick has Bag B. They draw with replacement from their bags at the same time (independently) until someone draws a white ball. (Then the person who did not draw a white ball does the dishes.) Find the probability that...

a. they draw the first white ball at the same time (after equal number of attempts). Round your answer to 4 decimal places.

b. Dick draws the first white ball before Jane does. Round your answer to 4 decimal places.

I know how a joint distribution works, but don't know how to apply it to this word problem, any tips?

Thanks for the help
• Topic Stats
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Type: Question • Score: 0 • Views: 2,478 • Replies: 6

fresco

2
Mon 13 Mar, 2017 03:31 am
@Rifscape,
It does not look like a distribution problem to me...merely the probability of a joint event (like totalling the score on two dice)
Use a two dimensional sample space (with A outcomes across by B outcomes down) You are then merely counting squares for the joint event and dividing by the total number of squares(7x15).
or
a. multiply the probabilities for both white.
b. multiply the probabilities of (Dick white) x ( NOT Jane white)
0 Replies

engineer

3
Mon 13 Mar, 2017 07:20 am
@Rifscape,
Here are the basic scenarios.

Both Black draws = 6/7 * 14/15 = 4/5
Dick White, Jane Black = 1/15 * 6/7 = 2/35
Dick Black, Jane White = 1/7 * 14/15 = 2/15
Both White draws = 1/7 * 1/15 = 1/105

This is a recursive problem. If both are black, we replace the balls and draws again, getting the exact same probabilities, so let's look at Dick drawing white and Jane drawing black. The chance is the sum of the first draw probability (2/35), plus the second draw probability (4/5 * 2/35) plus the third draw probability (4/5)^2 * 2/35 ... This is a geometric series of the form

2/35 * (1 + 4/5 + (4/5)^ + (4/5)^3 + ... ) = (2/35) / (1 - 4/5) = 2/7.

Using the same technique, the probability of Dick drawing black and Jane drawing white is 2/3 and the probability of both drawing white is 1/21.
ekename

1
Mon 13 Mar, 2017 09:32 pm
@engineer,
On the other hand , if you've had too much

Fun With Dick and Jane

and you're fast running out of time to do your other homework you could try the

See Spot Run Solution:

Jane has 6 chances of losing and 1 chance of a tie.
Dick has 14 chances of losing and 1 chance of a tie.
The cumulative probability is therefore:
1 chance in 21 of a tie = 1/21
14 chances in 21 of Jane winning = 2/3
6 chances in 21 of Dick winning = 2/7
inkhumming

1
Tue 14 Mar, 2017 04:19 am
@ekename,
I think you just made that up and it isn't a solution to the problem.
ascribbler

1
Tue 14 Mar, 2017 04:26 am
@inkhumming,
I agree. I suspect ekename is probably half in the bag with all the silliness that it/she/he spouts about everything in general and math in particular.

Quote:
Bag A has 6 black balls and 1 white ball. Bag B has 14 black balls and 1 white ball. Jane has Bag A and Dick has Bag B.
0 Replies

ekename

2
Wed 15 Mar, 2017 06:48 am
@engineer,
Engineer, do you like my solution or do you think it's a teensy bit

0 Replies

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