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Vegita Thu 29 Oct 2015 19:55:23 No.7624760 View Reply Report Delete

A box contains 10 tickets with the numbers
{1,1,1,2,2,2,2,3,3,4}
.If tickets are drawn successively
without replacement, what is the probability that it takes exactly 5 draws to see the first 4?

Anonymous

Anonymous Thu 29 Oct 2015 19:57:58 No.7624769 Report Delete

Quoted By:

>>7624760
1/10x 1/9x 1/8x 1/7x 1/6x 1/5

Anonymous

Anonymous Thu 29 Oct 2015 19:59:23 No.7624774 Report Delete

Quoted By:

P
(exactly 5 draws to see the first 4) =
P
(no 4 in first four draws, 4 on the 5th
draw) =
P
(no 4 in first four draws)
·
P
(4 on 5th draw
|
no 4 in first four draws)
.
The number of 4s in the first four draws has a Hypergeometric(4,10,1) distribution,
and the conditional probability of a 4 on the 5th draw given no 4s in the first four
draws is
1
6
(one out of the 6 tickets remaining is a 4). Therefore, the probability is

Anonymous

Anonymous Thu 29 Oct 2015 20:01:16 No.7624783 Report Delete

Quoted By: >>7624798

>>7624760

P(no,10)P(no,9)P(no,8)P(no,7)P(yes,6)
9/10*8/9*7/8*6/7*1/6
1/10

Anonymous

Anonymous Thu 29 Oct 2015 20:05:59 No.7624798 Report Delete

Quoted By:

>>7624783
It's equally likely that the 4 ticket will be picked on any one of the 10 draws so on the fifth it will be 1/10

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