Question

Of the two officers selected from a group of officers, the probability of the first being male and the second being female is 1/4. Since the ratio of male civil servants to female civil servants is 1/2, how many civil servants are there in the group? b) One bag contains 6 white and 4 black balls. What is the probability that one of the 3 balls drawn randomly from this bag is white and the other two are black?

Answer 1

1)

Let X is number of males

Let Y is number of females

We are given :

X/Y = 1/2

=> Y = 2X

Probability that first is male and second is female = 1/4

Total servants = X + Y

Probability that first is male = X / (X+Y)

Now, one male is taken out so there are total X+Y -1 servants remaining.

Probability that second is female given first was male = Y /( X+Y-1)

So,

Probability that first is male * Probability that second is female given first was male = Probability that first is male and second is female

Substituting Y = 2X

Since, Y = 2X

So, Y = 2*3 = 6

Total civil servants in the group = X + Y = 3 + 6 = 9

2)

There are in total 10 balls : 6 white and 4 black

Out of 10 , 3 are selected

In this , out of 6 white only 1 is selected and out of 4 black only 2 are selected

Probability that one of the 3 balls drawn randomly from this bag is white and the other two are black =

Note : n! = n*(n-1)*(n-2)*.........................*1

Solving this we get :

Probability that one of the 3 balls drawn randomly from this bag is white and the other two are black = 0.3