Probability help!!! With steps please

Statistics
Tutor: None Selected Time limit: 1 Day

A Jar has 20 marbles in it, 5 of which are yellow, the rest green. You are selecting 6 at a time. What is the probability that:

  you get exactly 1 yellow?   Probability at least 2 are yellow?

May 12th, 2015

This is a binomial distribution with P(S) = 5/20 =1/4 of choosing a yellow marble.

Six out of 20 are chosen so  number of combinations

                                                          = (20 x 19 x 18 x 17 x 16  x 15) x 1/ 6!

=3870

More time needed will take another 20 mins

May 12th, 2015

P(first marble is yellow, remaining 5 are green) = 5/20 x P( choosing sequence of 5 green out of 19 without replacement)

= 1/4 x 15/19 x 14/18 x 13/17 x 12/16 x 11/15 

= 360360/1395360

But yellow could come in 6 different ways

P(one yellow) = 1001/6460

Second part - doing

                                

May 12th, 2015

P(0 yellow) = P(sequence of 6  greens from 20 without replacement)

= 15/20 x 14/19 x 13/18 x 12/17 x 11/16 x 10/15

=3603600/27907200 

=1001/7752

Hence P(of 0 or 1 yellow) = 1001/66460 + 1001/7752

=    11011/38760

Hence P( at least 2 yellow) = 1 - P(0 or 1 yellow) 

=1- 11011/38760

= 27749/38760





May 12th, 2015

Number of ways of choosing any 6 from 20 = 20 C 6 = Six out of 20 are chosen so  number of combinations

 = (20 x 19 x 18 x 17 x 16  x 15) x 1/ 6!

=38760

Number containing 5 greens and 1 yellow from these = 

Number of ways of choosing 5 greens from 15 and 1 yellow from 5 yellows

= 15 C5 x 5

= (15 x 14 x 13 x12 x 11/ 5!) x 5

= 15015

Hence P(one yellow) = 15015/38760 = 1001/2584


May 12th, 2015

Final please ignore all others

Part 1

Number of ways of choosing any 6 from 20 = 20 C 6 = Six out of 20 are chosen so  number of combinations

 = (20 x 19 x 18 x 17 x 16  x 15) x 1/ 6!

=38760

Number containing 5 greens and 1 yellow from these = 

Number of ways of choosing 5 greens from 15 and 1 yellow from 5 yellows

= 15 C5 x 5

= (15 x 14 x 13 x12 x 11/ 5!) x 5

= 15015

Hence P(one yellow) = 15015/38760 = 1001/2584


Part 2

Second part

Number containing 0 yellows = 15 C 6 =  15 x 14 x 13 x 12 x 11 x 10  x 1/6! = 5005

Hence P( 0 yellows) = 5005/38760 = 1001/7752

Hence P(at least 2 yellows) = 1- 1001/2584 – 1001/ 7752

1- 4004/7752

= 3748/7752

= 937/ 1938




May 12th, 2015

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