Tough probability question

Question

Two numbers are picked from 1st 20 natural numbers. What is the probability of getting the 1st number a multiple of 3 and second number a multiple of 2.

57/380

Note: I got this from Manhattan review class and cannot seem to get the answer provided. Any help on this would be appreciated.

maliyeci · Answer

There are two cases. 
1st, two numbers can be equal. In this case
probability is 6/20x10/20=60/400=3/20
2nd two numbers cannot be equal
in this case we will add two different probabilities
3/20x10/19+3/20x9/19=57/380

anandxrfeu · Answer

Thank you for your reply.

Should we not be considering other mutually exclusive cases such as the following:
  case a : first number multiple of 3 but not a multiple of 2  AND  second number a multiple of 2 but not a multiple of 3.
  case b:  first number multiple of 3 but not a multiple of 2  AND  second number a multiple of 2 which is also a multiple of 3. 
  case c : first number multiple of 3 which is also a multiple of 2  AND  second number a multiple of 2 but not a multiple of 3.
  case d:  first number multiple of 3 which is also a multiple of 2  AND  second number a multiple of 2 which is also a multiple of 3.

on solving :
 case a: 3/20 x 7/20
 case b: 3/20 x 3/20
 case c: 3/20 x 7/20
 case d: 3/20 x 3/20

Ans: case a + case b + case c + case d
 : 3/20

Where am i going wrong?

maliyeci · Answer

1st-Your second denominator is wrong, because you are selecting from 19 not 20.
2nd in case d you are grouping
6-12-18 with 6-12-18 and taking 9 possibilities but there are only 6. It makes 60/380-3/380=57/380

anandxrfeu · Answer

You said my 2nd denominator is wrong. But the question doesnt mention that we can not select the the same number again does it?

maliyeci · Answer

If you look at my first answer, you can see that, I put two cases about the question. Since it is not an official question, there may be some flaws like that. I say that your denominator is wrong, because I found that from the OA that, the question does not allow to select the same number.

anandxrfeu · Answer

Got it <img src="{SMILIES_PATH}/icon_smile.gif" alt=":)" title="Smile" />

Postal · Answer

But why do we count 9/19 in the second case??

shouldn't we count 3/20 * 10/19 + 10/20 * 3/19 ??

maliyeci · Answer

There are shared numbers such as 6,12,18. Thus, if you select a number which can both divisible by 2 and 3, you take one element from the set of numbers that are divisible by 2.

mydreammba · Answer

1st case is right but in second case how did you tak3 3/20 i believe it should be

6/20x10/19

Can you plz explain more clearly if i am wrong?

maliyeci · Answer

There are 3 numbers which are divisible by 3. 3,6,9,12,15,18. 3,9 and 15 are not divisible by 2 so 3/20x10/19 comes here. 6, 12 and 18 are divisible both by 2 and 3 so if you select one of them there will remain only 9 numbers that can be divisible by 2. That makes 3/20x9/19 OK?

shahideh · Answer

Multiples of 3 are: 3-6-9-12-15-18
we see that 3 of them are even and therefore, multiple of 2. So, we should break them:

P=(probability that first is multiple of 3 but not 2, and second is multiple of 2) + (probability that first is multiple of 3 and 2, and second is multiple of 2)

P= \frac{3}{20}*\frac{10}{19} +\frac{3}{20}*\frac{9}{19}

Notice that in above equation, 3 in the first  \frac{3}{20}  is representative of 3 and 9 and 15, while 3 in the second  \frac{3}{20}  is representative of 6 and 12 and 18.

P=\frac{3}{20}

devinawilliam83 · Answer

I think the ques is ambiguous, we can assume either the number is replaced after the 1st pick (in which case we will have repeat nos.) or the number is not replaced, in which case the answer is 51/380

Tough probability question

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