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  • A bag contains 8 red and 5 white balls. Three balls are drawn at random. Find the Probability that: (a) All the three balls are white (b) All the three balls are red (c) One ball is red and two balls are white

A bag contains 8 red and 5 white balls. Three balls are drawn at random. Find the Probability that:
(a) All the three balls are white
(b) All the three balls are red
(c) One ball is red and two balls are white

Answers (1)

Given:

No. of reds balls = 8

No. of white balls = 5

Thus, total no. of balls, n = 13

Now, 3 balls are drawn at random, thus,

r = 3

Thus,

n (S) = nCr

            = 13C3

  1. If all balls are white,

P (A) = n (A)/ n(S)

      = no. of favorable outcomes/ sample space

Now, total white balls are = 5

Thus,

P (all the three balls are white) = 5C3 / 13C3

                  =\frac{\frac{5! }{3!\left ( 5-3 \right )}}{\frac{13! }{3!\left ( 13-3 \right )}}

                 =\frac{\frac{5\times 4\times 3! }{3!2!}}{\frac{13\times 12\times 11\times 10! }{3 \times 2 \times 1\times 10!}}

                 =\frac{\frac{5\times 4 }{2\times 1}}{13\times 2\times 11}

                      = 5 /143

  1. All three balls are red,

P (A) = n (A)/ n(S)

      = no. of favorable outcomes/ sample space

Now, total red balls are = 8

Thus,

P (all the three balls are white) = 8C3 / 13C3

                

  =\frac{\frac{8! }{3!\left ( 8-3 \right )}}{\frac{13! }{3!\left ( 13-3 \right )}}

                 =\frac{\frac{8\times 7\times 6\times 3! }{3!5!}}{\frac{13\times 12\times 11\times 10! }{3 \times 2 \times 1\times 10!}}

                 =\frac{\frac{8\times 7\times 6 }{2\times 3}}{13\times 2\times 11}

                = 28 /143

  1. One ball is red and two balls are white

P (A) = n (A)/ n(S)

      = no. of favorable outcomes/ sample space

Now, total white balls are = 5

Thus,

P (One ball is red and two balls are white) = 8C1 x 5C2 / 13C3

                 

  =\frac{\frac{5! }{3!\left ( 5-2 \right )}\times \frac{8! }{3!\left ( 8-3 \right )}}{\frac{13! }{3!\left ( 13-3 \right )}}

                 =\frac{\frac{8!\times 7! }{7!}\times \frac{5\times 4\times 3! }{2\times 1\times 3!}}{\frac{13\times 12\times 11\times 10! }{3 \times 2 \times 1\times 10!}}

                 =\frac{8\times 10 }{13\times 2\times 11}

                = 40 /143

Posted by

infoexpert22

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