A batch consists of 12 defective coils and 88 good ones. find the probability of getting two good coils when two coils are randomly selected if the first selection is replaced before the second is made.
Help me..
since it would be replaced would the answer be 0.7744 ?
7 answers
No. (76/88)^2 = 0.7459
how is it going to be 76/88 ?
It isn't. For two draws, it is the square of that.
There are 76 good coils out of 88. What does that tell you about the probabiloity of selecting a good one at random each time?
There are 76 good coils out of 88. What does that tell you about the probabiloity of selecting a good one at random each time?
My answer was correct.. Your answer is not in the choices :S
There are a total of 100 coils, of which 88 are good
prob of getting 2 good ones in a row, after replacing the first
= (88/100)(88/100)
= (88/100)^2
= .7744
prob of getting 2 good ones in a row, after replacing the first
= (88/100)(88/100)
= (88/100)^2
= .7744
Thank you Reiny, that is what I thought.
Drwls I appreciate your effort anyway :)
Drwls I appreciate your effort anyway :)
a batch consists of 12 defective clocks and 88 good ones find the probability of getting two good clocks when the two are randomly selected if the first selection is replaced before the second selection is made.
B- the first selection is not replaced.
B- the first selection is not replaced.
1 answer