- #1
veronica1999
- 61
- 0
Forty slips are placed into a hat, each bearing a number 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10 with each number entered on four slips. Four slips are drawn from the hat at random and without replacement. Let http://www.artofproblemsolving.com/Forum/code.php?hash=516b9783fca517eecbd1d064da2d165310b19759&sid=c2203dbdfd142396d8e0d12f81599d60 be the probability that all four slips bear the same number. Let http://www.artofproblemsolving.com/Forum/code.php?hash=22ea1c649c82946aa6e479e1ffd321e4a318b1b0&sid=c2203dbdfd142396d8e0d12f81599d60 be the probability that two of the slips bear a number http://www.artofproblemsolving.com/Forum/code.php?hash=86f7e437faa5a7fce15d1ddcb9eaeaea377667b8&sid=c2203dbdfd142396d8e0d12f81599d60 and the other two bear a number http://www.artofproblemsolving.com/Forum/code.php?hash=d6bdb677e4af59f78f088681f60ac28a0d6b031b&sid=c2203dbdfd142396d8e0d12f81599d60. What is the value of http://www.artofproblemsolving.com/Forum/code.php?hash=bf91cc5da690d4285591cb473adfc911d2794d13&sid=c2203dbdfd142396d8e0d12f81599d60?
For p there are only 10 cases when all the numbers are the same.
For q there would be a lot more cases.
First, there will be 10C2 ways to choose the two numbers then i have to think of the ways the number can be arranged. xxyy can be arranged in 6 ways and then i have to multiply by two because the number x and y can be switched.
So my answer is 10C2X6X2/40C4 over 10/40C4
I am not sure why my answer is wrong.(Worried)
For p there are only 10 cases when all the numbers are the same.
For q there would be a lot more cases.
First, there will be 10C2 ways to choose the two numbers then i have to think of the ways the number can be arranged. xxyy can be arranged in 6 ways and then i have to multiply by two because the number x and y can be switched.
So my answer is 10C2X6X2/40C4 over 10/40C4
I am not sure why my answer is wrong.(Worried)