Question

By accident, 8 burned out bulbs have additionally been mixed in with 30 good ones. Ken is replacing old bulbs in his house. If he selects two bulbs at random from the box of 38, what is the probability they both work?
P(Both Work) =
435 / 703


P(Both Work) =
28 / 703


P(Both Work) =
16 / 361


P(Both Work) =
169 / 256

Answer 1

Answer:

The correct answer option is P(Both Work) =  435 / 703

Step-by-step explanation:

We are given that in a box of total 38 bulbs, 8 are the defective ones while the rest are good.

If Ken selects two bulbs at random from the box, we are to find the probability that they both work.

1st bulb: P (bulb works) =  $\frac{30}{38}$

2nd bulb: P (bulb works) = $\frac{29}{37}$

P (Both Work) = $\frac{30}{38} \times \frac{29}{37}$ = 435 / 703

Answer 2

Answer:

The correct answer is first option.   P(Both Work) =  435/703

Step-by-step explanation:

It is given that,

There are 8 burned bulbs and 30 good bulbs.

Therefore total number of bulbs = 8 + 30 = 38

To find the probability

We have to take two working bulbs from total bulbs.

Total bulbs = 38

The probability of taking two working bulbs = P(Both Work)

P(Both Work) = 30C₂/38C₂

  = (30 * 29)/(38 * 37) = 435/703

Therefore the correct answer is first option