SOLUTION: Indepent Events I'm having a hard time understanding this please can someone explain in detail. A bag contains 8 apples, 5 are eating apples and 3 are cooking apples. If 2 ap

Algebra ->  Finance -> SOLUTION: Indepent Events I'm having a hard time understanding this please can someone explain in detail. A bag contains 8 apples, 5 are eating apples and 3 are cooking apples. If 2 ap      Log On


   

Question 1068589: Indepent Events
I'm having a hard time understanding this please can someone explain in detail.

A bag contains 8 apples, 5 are eating apples and 3 are cooking apples. If 2 apples are drawn without replacement.Find the probability that
A) the first apple is an eating apple and the second apple is a cooking apple
B)one of the apples are an eating apple
Found 2 solutions by stanbon, KMST:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A bag contains 8 apples, 5 are eating apples and 3 are cooking apples.
If 2 apples are drawn without replacement.
Find the probability that
A) the first apple is an eating apple and the second apple is a cooking apple
Ans: (5/8)(3/7) = 15/56
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B)one of the apples is an eating apple.
I'll assume you mean "exactly one of the apples is an eating apple".
Ans: (5/8)(3/7)+(3/8)(5/7) = (15/56)+(15/56) = 30/56
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Cheers,
Stan H.
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Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
The drawing of the first apple changes the ratio of the two types of apples.
That ratio is not the same for the picking of the second apple.

A) For the first pick the probability of getting an eating apple is
5%2F%285%2B3%29=5%2F8 .
That means that if you did it a million times 5%2F8 of the first apple picks would be eating apples.
For the second pick there is a total of 8-1=7 apples,
and if the first pick was an eating apple,
there are still 3 cooking apples in the bag,
so your chance of picking a cooking apple after having picked an eating apple is
3%2F7 .
All in all, if you did the experiment a million times,
5%2F8 of those times your first pick would be an eating apple,
and of those times
3%2F7 of the time your second pick would be a cooking apple,
so that sequence of events would happen %285%2F8%29%2A%283%2F7%29=highlight%2815%2F56%29 of the times.

B) The wording you posted is suspicious.
I wonder if you may have been trying to type that neither one of the apples picked is an eating apple.
There is a 15%2F56 probability that a cooking apple is picked after an eating apple is picked.
There is a %283%2F8%29%2A%285%2F7%29=15%2F56 probability that a cooking apple is picked before an eating apple is picked.
There is a %283%2F8%29%282%2F7%29=6%2F56 probability that first pick and the second pick are both cooking apples.
There is a %285%2F8%29%284%2F7%29=20%2F56 probability that first pick and the second pick are both eating apples.
What are we asked for?
The probability that none of the two apples is an eating apple? That is 6%2F56 .
The probability that exactly one of the apples picked is an eating apple? That is 15%2F56%2B15%2F56=30%2F56 .
The probability that at least one of the two apples picked is an eating apple (and maybe both are)>? That is 1-6%2F56=50%2F56 .