Conditional Probability question - Light bulbs

[megcode]

Conditional Probability question:

25% of the bulbs in a box are damaged. The first 2 bulbs are not damaged.

What is the probability that the bulb picked on the third attempt is damaged?

Ans: 9/64

Can someone help explain, how the answer is derived?

 

[Deleted member 4993]

Conditional Probability question:

25% of the bulbs in a box are damaged. The first 2 bulbs are not damaged.

What is the probability that the bulb picked on the third attempt is damaged?

Ans: 9/64

Can someone help explain, how the answer is derived?

Probability of first bulb NOT being damaged = 0.75

Probability of (first+second) bulb NOT being damaged = 0.75*0.75

..............................continue

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[megcode]

I did 75/100 * 75/100 * 25/100 = 9/64

Do you have any good recommendations for teaching probability for Kids 6th to 12th grade?

Thank you.

 

[blamocur]

Conditional Probability question:

25% of the bulbs in a box are damaged. The first 2 bulbs are not damaged.

What is the probability that the bulb picked on the third attempt is damaged?

Ans: 9/64

Can someone help explain, how the answer is derived?

Does not this probability depend on the total number of the bulbs in a box? I would think that for a box with 4 bulbs the probability would be 1/2, but for 8 bulbs my answer would be 1/3.

 

[Dr.Peterson]

I did 75/100 * 75/100 * 25/100 = 9/64

No, that would be the probability that the first two are good AND the third is bad, assuming the box is large.

The probability that the third is bad is 25%, under the same assumption! The problem did not ask for the probability of all three bulbs being in the indicated states, but just the third (given the information about the first two), which is a conditional probability just as your title says:

Conditional Probability question:

25% of the bulbs in a box are damaged. The first 2 bulbs are not damaged.

What is the probability that the bulb picked on the third attempt is damaged?

Ans: 9/64

Who said 9/64 was the correct answer?

If, on the other hand, it is a box of four, then one is bad, which is 1/2 of those remaining, as blamocur said. This, again, is the conditional probability. And if it's a box of 8, then 2 are bad, out of the 6 remaining, for a probability of 1/3.

 

[megcode]

No, that would be the probability that the first two are good AND the third is bad, assuming the box is large.

The probability that the third is bad is 25%, under the same assumption! The problem did not ask for the probability of all three bulbs being in the indicated states, but just the third (given the information about the first two), which is a conditional probability just as your title says:

Who said 9/64 was the correct answer?

If, on the other hand, it is a box of four, then one is bad, which is 1/2 of those remaining, as blamocur said. This, again, is the conditional probability. And if it's a box of 8, then 2 are bad, out of the 6 remaining, for a probability of 1/3.

The test software mentioned 9/24 is correct - i agree it is combined probability of all 3

If it is only third being bad it is 25/100 or 1/4