198 700+ problem question 16

Question

"A basket has 5 apples, one is spoiled. If Henry picked 2 apples simultaneously and at random, what is the probability that the 2 selected apples will include the spoiled one.

a) 1/5
b) 2/10
c) 2/5
d) 1/2
e) 3/5

Hi guys,

I'm kinda stuck on the reasoning for the answer being "C". Intuitively I thought 2/5 because in my reasoning, picking 2 simultaneously doesn't change the fact that 1/5 apples is always spoiled, and the remaining apple that can be picked to make up the 2 is 1/5?

Not sure if my reasoning makes sense. Could some please help to provide the explanation for this question? Thanks much in advance!

asamaverick · Answer

Here is how I would approach this.

Picking 2 apples from the basket containing 5 : You can select this in 5C2 = 10 ways.
Out of these 10 ways, there will be 4 instances in which the spoiled one is selected.
If you denote the apples as G1, G2, G3, G3 & S. The four combinations of picking spoiled one are SG1, SG2, SG3 & SG4. (Remember order does not matter since we are picking them simultaneously SG1 = G1S).

So the probability = 4/10 = 2/5.

jube · Answer

Another approach:

Probability that one of the apples will be spoiled = 1- Probability that neither of the apples are spolit

Probability that neither of the apples are spoilt: (4/5)(3/4)=3/5

Answer: 1-(3/5)=2/5

kvcpk · Answer

I am not sure if my approach is rite wor wrong.. I want an expert to correct me..

The question is same as picking 1 apple each time for 2 times.. but with an OR condition.

So the first time when i pick, probability to pick the spoilt apple is 1/5.
second time to pick it is also 1/5 [from the same apple set again]

we know that one of these two should occur. So 1/5 + 1/5 = 2/5

Please let me know if i am wrong!!

indiantiger · Answer

the probability of getting a spoiled apple is 1/5

we can have total two cases :

Case 1 first one is spoil and second is fine.

Case 2 first one is fine and second one is spoil

so total cases * probability of the event

2 * 1/5 = 2/5

Please correct me if I am wrong.

sk818020 · Answer

I solved it rather quickly by:

Probability you the first one you draw is poisoned, a = 1/5
Probability the second is not poisoned, b = 4/4

1/5*4/4=4/20

Probability the first you draw is not poisoned, c = 4/5
Probability the second drawn is poisoned, d = 1/4

4/5*1/4=4/20

4/20+4/20=8/20=2/5

Or using statistical notation;

[P(a)*P(b)]+[P(c)*P(d)]=4/20+4/20=2/5

Hope that helps.

Thanks,

Jared

artistocrat · Answer

I really like some of the methods for solving this problem that are described here. Here is another way to look at the problem.

So you have five apples: 1,2,3,4,5 (let's say 1 is rotten)

You have 10 pairs (taking out 2 "simultaneously"):

12 23 34 45
13 24 35
14 25
15

Notice that 1 is found in 4 pairs of 10.

4/10=2/5

artistocrat · Answer

I really like some of the methods for solving this problem that are described here. Here is another way to look at the problem.

So you have five apples: 1,2,3,4,5 (let's say 1 is rotten)

You have 10 pairs (taking out 2 "simultaneously"):

12 23 34 45
13 24 35
14 25
15

Notice that 1 is found in 4 pairs of 10.

4/10=2/5

Psyker · Answer

Thanks much to everyone for their help! really useful to see the different approaches.

kvcpk · Answer

kvcpk wrote:I am not sure if my approach is rite wor wrong.. I want an expert to correct me..

The question is same as picking 1 apple each time for 2 times.. but with an OR condition.

So the first time when i pick, probability to pick the spoilt apple is 1/5.
second time to pick it is also 1/5 [from the same apple set again]

we know that one of these two should occur. So 1/5 + 1/5 = 2/5

Please let me know if i am wrong!!

Can some expert confirm if my approach above is correct?

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