Question

1.Two (2) balls are drawn in succession without replacement from a box containing four (4) blue balls and five (5) green balls. The possible outcomes and the value z of the random variable Z, where Z is the number of blue balls are shown in the table:

Answer 1

Box contains 4 blue & 5 green balls. Four balls are are taken out randomly from the box.The total no.of ways in which this can be done is equal to C(9,4) = 9!/(4!×5!) =126. Now the possible combination of 4 drawn balls (2b,2g) are as follows ;

C(4,2)×C(5,2) =(4!/2!×2!)×(5!/3!×2!) =6×10=60.

Therefore required probability = 60/126 =10/21.

Note that C(n,r) = {n!/r! ×(n-r)! }.