### Problem 5 - Pick the higher number

There are 15 folded chits numbered 1 to 15. Aish comes and picks a chit, notes the number and puts it back into the lot. Abhishek comes and picks a chit from the lot. What is the probability that the chit Abhishek has picked up has number higher than the one picked by Aishwarya.

The rule as always is to 'think simple' :)

The rule as always is to 'think simple' :)

## 4 Comments:

my shot at it :7/15

at any pt the prob of A picking a chit is 1/15.

now if A has 15, cases that B>A ==0. Prob = 0/15, 15 = totsl no. of possiblities

A 14 , cases that B>A ==1 , Prob = 1/15

anad so on till 14/15..

Thus for both events to occur simultaniously multiply prob and sum it == 7/15

My answer .

the probablity of Abhi picking a number bigger than Aish is 7/15.

Total no. of possible chit combinations are 225 (15 x 15 ) . If Aish picks 1 then from Abhi's choices 14 are favourable to our event (2 to 15) . Similarly if she picks 2 , then 13 of his choices (3 to 15 ) are favourable to event . . . and so on . Thus the total number of cases favourable to event are 14 +13 +12. . .+1 i.e. 105 . Hence the probability of our event is 105/225 or 7/15 .

yeah the answer is 7/15.

same logic.

total combinations= 15*15.

odds for abhishek picking higher no = 14+13+..+1=105

ans= 105/225

Guys,

You hit it right. My only input on this would be to arrive at the same solution in a simpler way with less calculation.

The way you have gone about it is using conditional probability. As I have said in some of the previous problems, conditional probability could involve some calculations and in some cases this could be avoided. (remember the hidden balls problem). But note that all conditional probability problems may not have a simpler way out.

So the solution:

1. Say Aish picks up a number X.

2. The probability of Abhi picking up the same X is 1/15

3. Apart from the above case, where it could be equal, the chances of Abhi picking a higher number is equal to the chances of Abhi picking a lower number.

4. Hence the probability = (1-1/15)/2 = 7/15

Think about how this came about to be equal to your answer.

You said p = (n*(n+1)/2) /(15*15) where n = 14

The learning from this is, when there are a set of cases which are mutually exclusive, equally likely and form a comprehensive set, there is a high probability that you have a simpler way of getting the answer than going through conditional probability.

Post a Comment

<< Home