Answered

A box contains 20 lightbulbs, of which 5 are defective.

If 4 lightbulbs are picked from the box randomly, the probability that at most 2 of them are defective is

a.31/969

b.70/323

c.91/323

d.938/969

Verified Answer :

Answer:

The correct option is d.

Step-by-step explanation:

It is given that the box contains 20 light bulbs, of which 5 are defective.

The possible ways of selecting r items from total n.

\(^nC_r=\frac{n!}{r!(n-r)!}\)

The total possible ways to select 4 light bulbs form 20 bulbs is

\(^{20}C_4=4845\)

We have to find the probability that at most 2 of them are defective.

Total possibility of selecting at most 2 defective bulbs.

Total = 0 bulb is defective + 1 bulb is defective +2 bulb is defective

\(Possibility=^{15}C_4\times ^{5}C_0+^{15}C_3\times ^{5}C_1+^{15}C_2\times ^{5}C_2\)

\(Possibility=1365\times 1+455\times 5+105\times 10=4690\)

The probability of selecting at most 2 defective bulbs is

\(P=\frac{\text{Required possibility}}{\text{Total possibility}}\)

\(P=\frac{4690}{4845}\)

Divide the numerator and denominator by 5.

\(P=\frac{938}{969}\)

The correct option is d.

yorkpeppermint

Answer:

Step-by-step explanation:

938/969 is right just took test got a 5/5

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