six burned-out bulbs have been mixed in with 16 good ones. Ken is replacing old bulbs in his house. If he selects two bulbs at random from the box of 22, what is the probability they both work?

P(Both work) = 64/121
P(Both work) = 61/121
P(Both work) = 8/11
P(Both work) = 40/77

Question

six burned-out bulbs have been mixed in with 16 good ones. Ken is replacing old bulbs in his house. If he selects two bulbs at random from the box of 22, what is the probability they both work?

P(Both work) = 64/121
P(Both work) = 61/121
P(Both work) = 8/11
P(Both work) = 40/77

welshfella · Answer

probability of the first one picked being OK = 16/22 or 8/11

what do you think probability of 2nd one being ok is?

anonymous · Answer

15/21 ??

welshfella · Answer

exactly right 
which reduces to 5/7 right?

anonymous · Answer

yes, but why isnt it one of the possible answers that is why i am confused

anonymous · Answer

oh wait, i have to multiply them now right

welshfella · Answer

because you have 2 events  so you have either to add or multiply the probabilities.
In this case we multiply them because the 2 'picks' are independent of each other.

so final answer is 3/11 * 5/7

anonymous · Answer

i see

welshfella · Answer

good

anonymous · Answer

8/11 x 5/7 = 40/77  answer is D

anonymous · Answer

thank you!

welshfella · Answer

correct 

a little more than 50% 

yw