Question

Three numbers are chosen at random without replacement from {1, 2, 3, ...... 8}. The probability that their minimum is 3, given that their maximum is 6, is

Answer 1

Answer:

0.2

Step-by-step explanation:

In general, the total number of ways of choosing r numbers out of given n numbers are nCr.

Let three numbers be l(minimum or least) ,m(middle) and h(maximum).  

Given the maximum of 3 is 6, i.e., h = 6, thus the other two are less than 6.Therefore, the total number of ways of choosing other two numbers (l,m) are 5C2 = 10.

Hence, total number of cases are = 10.  

If minimum or least(l) = 3 and also h =6 , so m is in between 3 and 6 .e., m could be 4 or 5. Thus there are 2 favourable cases.  

Thus, the probability = (Total number of favourable cases)/(Total number of cases) = $\frac{2}{10}$ = 0.2.