Question

A box contains 4 White marbles and 5 blue marbles. What is the probability of drawing 2 blue marbles and 1 White marble in succession without replacement?

Answer 1

no. of white marbles= 4

no. of black marbles= 5

total= 9

(since it is the case of without replacement, the sample space will keep on decreasing after each draw)

probability of drawing blue marble in 1st draw= $\frac{5}{9}$

probability of drawing blue marble in 2nd draw= $\frac{4}{8}$

probability of drawing white marble in 3rd draw= $\frac{4}{7}$

hence, Required probability = $\frac{5}{9}$ × $\frac{4}{8}$ × $\frac{4}{7}$

                                               = $\frac{10}{63}$

Answer 2

Given,

Number of white marbles = 4

Number of blue marbles = 5

To Find,

The probability of drawing 2 blue marbles and 1 White marble in succession without replacement =?

Solution,

Total marbles = 4 + 5 = 9

The probability of drawing blue marble 1st time = 5 / 9

Now, 1 blue marble is taken out and is not replaced. Total marbles = 8 and blue marbles = 4

The probability of drawing blue marble 2st time =  4 / 8

Similarly, the probability of drawing white marble for 3rd time =  4 / 7

The probability of drawing 2 blue marbles and 1 White marble in succession without replacement = (5/9)*(4/8)*(4/7)

The probability of drawing 2 blue marbles and 1 White marble in succession without replacement =  80 / 504

The probability of drawing 2 blue marbles and 1 White marble in succession without replacement  = 10/63

Hence, the probability of drawing 2 blue marbles and 1 White marble in succession without replacement is 10/63.