+67 Forty slips are placed into a hat, each bearing a number 1,2,3,4,5,6,7,8,9 or10 with each number entered on four slips. Four slips are drawn from the hat at random and without replacement. Let p be the probability that all four slips bear the same number. Let q be the probability that two of the slips bear a number a and the other two bear a number b doesnot equal to a. What is the value of q/p?
+67 The total number of ways that the numbers can be chosen is 40c4. Exactly 10 of these possibilities result in the four slips having the same number.
Now we need to determine the number of ways that two slips can have a number a and the other two slips have a number b , with \($b\ne a$\) . There are 10c2 ways to choose the distinct numbers a and b . For each value of a , there are 10c2 ways to choose the two slips with a and for each value of b , there are 4C2 ways to choose the two slips with b . Hence the number of ways that two slips have some number and the other two slips have some distinct numberb is
(10C2)(4c2)(4c2)=45*6*6=1620
So the probabilities p and q are 10/40c4 and 1620/40c4 , respectively, which implies that
q/p=1620/10=162
