A basket contains 5 apples of which 1 is spoiled and rest is

[Guest]

Source : Gmat prep 1

A basket contains 5 apples of which 1 is spoiled and rest is good. If henry is to select 2 apples from the basket simultaneously and at random , what is probability that 2 apples selected will include the spoiled apple?
1) 1/5
2) 3/10
3) 2/5
4) 1/2
5) 3/5

Ans- 2/5

[Guest]

I have an exam next week...So i will appreciate the explanation to above question from any smart fella here:)

[Guest]

Total ways of selecting 2 apples from the lot of 5 ==> 5P2 = 5!/2!3! = 10 ways
Ways in which one apple will be the spoiled one = 1 spoiled + any one of other 4 = 4 ways
Thus required Probability = 4/10 = 2/5

[Guest]

or, figure out the probability of NOT choosing the rotten apple in either pick and subtract that from 1.

1st pick: 4/5
2nd pick: 3/4

(4/5) * (3/4) = 12/20 = 3/5

1 - 3/5 = 2/5

[RonPurewal]

there's also a clever lil way in which you can rephrase this problem so that its solution becomes trivial.

specifically:
the problem is just asking for the probability that the spoiled apple is one of henry's picks.
note that two things happen in this problem:
(1) henry picks two apples;
(2) one of the apples spoils.
note that NOTHING REQUIRES THAT (1) HAPPEN BEFORE (2).

indeed, the problem is greatly simplified if you treat the situation as though (2) happens before (1).

let's say that henry has already selected his 2 apples, and that one of the apples then spoils. the question in the problem, then, is equivalent to the following question:
what is the probability that the spoiled apple is one of the 2 selected by henry?
this is just (# of apples selected by henry) / (total # of apples), or 2/5.

the problems on which you can do this sort of thing are rare indeed, but it's always satisfying to find solutions that circumvent the need for tiresome calculations.

[RR]

What is the significance of the word simultaneously in the question ?
Going by the answers, it seems to imply one after another without replacement.
However, our knowledge of the English language tells us otherwise !

A doubt here : If the problem were modified for with replacement, what would the answer be ?
A basket contains 5 apples of which 1 is spoiled and rest is good. If henry is to select 2 apples from the basket one after the other (he replaces the first apple in the basket, before selecting the second one) and at random , what is probability that the apple(s) selected will include the spoiled apple?

[RonPurewal]

What is the significance of the word simultaneously in the question ?
Going by the answers, it seems to imply one after another without replacement.
However, our knowledge of the English language tells us otherwise !

since the 2 apples selected are distinct, it makes no difference whether the selections are considered as "simultaneous" or "sequential".
they are still 2 independent events, and the probabilistic calculations proceed accordingly.

in the same way, the probabilistic analysis of flipping three coins simultaneously is exactly the same as the analysis of flipping one coin three times in a row, or that of flipping three coins one after the other.

--

A doubt here : If the problem were modified for with replacement, what would the answer be ?
A basket contains 5 apples of which 1 is spoiled and rest is good. If henry is to select 2 apples from the basket one after the other (he replaces the first apple in the basket, before selecting the second one) and at random , what is probability that the apple(s) selected will include the spoiled apple?

because you can now get the spoiled apple both times if you're especially unlucky, this would become "what's the probability of selecting the spoiled apple at least once?"
to find that probability, the easiest route is to use the OPPOSITE event, which is a relatively simple event: you don't pick the spoiled apple. by contrast, to calculate the probability directly, you'd have to find the probability of 3 things: spoiled then unspoiled, unspoiled then spoiled, and, finally, spoiled twice in a row.
the probability of not selecting the spoiled apple is (4/5)(4/5), or 16/25.
therefore, the probability that you'll get the spoiled apple at least once is 1 - 16/25, or 9/25.

[RR]

Thank you Ron. That was really helpful.

[AmunaGmat]

so guys I see all your explanation but I am still confused about my own thinking. Why my way is wrong:
1. The chance of picking a spoilt apple the first time is 1/5
2. The chance of picking the spoilt apple the second time is 1/4
Because you pick the first apple and then the 2nd apple, which I think simulaneously does not matter in this question, thus 1/5*1/4= 1/20

I know it is wrong because of the answer lol, but why?

Cheers!!

[tim]

you've just calculated the probability of getting two spoiled apples in a row, but even that is problematic because there is only one spoiled apple! basically what you've really calculated is this:

suppose apples A and B are spoiled, whereas C,D,E are not. the probability of picking A and then B is 1/5*1/4=1/20. so you've calculated the probability of picking two spoiled apples in a specific order from a basket of 5 where 2 are spoiled. be sure to think through what your fractions mean when you do a problem like this, and you'll be more likely to spot your errors..

[nilendud]

Amunagmat's approach will work if we did the following:

case1 : Probabilty of getting the spoilt apple in the first pick: 1/5
case 2: Probability of getting the spoilt apple in the 2nd pick: 4/5 (probability of getting the good apple) * 1/4 (probabilty of getting the bad apple from the remaining)

4/5*1/4=1/5

Probabilty of selecting the spoilt apple: case1 + case2 = 2/5

[RonPurewal]

Amunagmat's approach will work if we did the following:

case1 : Probabilty of getting the spoilt apple in the first pick: 1/5
case 2: Probability of getting the spoilt apple in the 2nd pick: 4/5 (probability of getting the good apple) * 1/4 (probabilty of getting the bad apple from the remaining)

4/5*1/4=1/5

Probabilty of selecting the spoilt apple: case1 + case2 = 2/5

yep, you can do that, too.

[nisha.mada]

You can select 2 apples out of 5 in 5C2 ways = 10 ways. You can select one bad and 1 good apple in any order in 1C1x 4C1 = 4 ways. Hence probability = desired outcome/tot outcomes = 4/10 = 2/5

[tim]

very nice! that's a great way to do this one..