2 Probability questions: Two cards are drawn from a deck of 52 playing cards without replacment. Determine the probability that:

a) the second card is a club, given the first one is a club

b) The second card is red, given the first is a diamond.

Please show your work

THe answers are

a) 4/17

b) 25/51

The second Question is:

If you rearrange the letters in the word DISTRIBUTION, what is the probability that RIB will not be together?

The answer is 19/22

Please show your work.

2 Probability questions: Two cards are drawn from a deck of 52 playing cards without replacment. Determine the

The second question is a little tricky.

There are 12 letters, with the i repeated 3 times and the t repeated twice.

So total arrangements (unique) = 12! / 2! / 3! = 11!

Now considering the RIB as a unit letter, there are now 10 letters, with the i and t being double.

So arrangements = 10! / 2! / 2!

Prob of RIB being together in that order =

10! / 2! / 2! / 11!

= 1/44

Prob (not together) = 43/44.

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NOW if you meant for the RIB to be together in ANY ORDER, then that greatly changes the problem.

You can just multiply the above result by 3! to get the arrangements of the letters R,I,B but there are some cases which will be overcounted. eg IRBI.

No of ways for R,I,B to be together in any order =

3! * 10! / 2! / 2!

This will give you the 6/44 or 3/22 you're looking for, but it's not correct yet.

Cases like IRBI and IBRI will each be counted twice. So we must subtract them.

No of ways IRBI occurs together = 9!/2!

The same thing for IBRI.

So we must subtract 9! from the above result.

No. of unique ways together in any order

= 3! * 10! / 2! / 2! - 9! = 5080320

Prob = 5080320/11! = 7/55

P(not together) = 48/55

Reply:the first is a club, means that there are 51 more cards, of which 12 are clubs

so 12/51 = 4/17

If the first is a diamond, then there are 51 more cards, of which 25 are red and 26 are black

so 25/51

______________________________________...

we look at the total number of ways to rearrange:

twelve letters, with repeats of I(3) and T(2)

12! / (3!2!) = 11!

of these RIB will be together (think of RIB as single letter)

so there are 10 items to be arranged with I(2) and T(2)

10! / (2!2!)

Do the division:

10! / 4

________

11!

= 1/44 will be together

????

Reply:I'm not in school anymore. Do your own work.

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