Question

A box contains 12 lamps of which 5 are defective.In how many ways can a sample of 6 be selected at random from the box so as to include atmost 3 defective sample

Answer 1

Answer:

805 ways

Explanation:

Total number of lamps = 12

Number of lamps defective = 5

A sample of 6 is to be picked (which includes at most 3 defective)

Sample of 6 with 1 defective lamp:

$5C_{1}$ x $7C_{5}$ = 5 x (7 x 6 x 5 x 4 x 3)/(5 x 4 x 3 x 2 x 1)

              = 105

Sample of 6 with 2 defective lamps:

$5C_{2}$ x $7C_{4}$ = [(5 x 4)/(2 x 1) x [(7 x 6 x 5 x 4)/(4 x 3 x 2 x 1)]

             = 350

Sample of 6 with 3 defective lamps:

$5C_{3}$ x $7C_{3}$ = [(5 x 4 x 3)/(3 x 2 x 1) x [(7 x 6 x 5)/(3 x 2 x 1)]

            = 350

Therefore, the number of ways in which a sample of 6 lamps can be selected with at the most 3 defective lamps = 105 + 350 + 350

                                                          = 805 ways