shreyashid wrote:
This is how I approached it. Please tell me if this is not right and why?
Total integers between 20-29=10
Number of prime =2 (23 & 29)
Number of integers divisible by 3 = 3 (21,24 & 27)
Total ways 2 integers can be picked from 10 integers : 10C2
Thus probability : 2*3/ 10C2 = 2/15
No. Note that the numbers are picked with replacement. 10C2 gives you the number of ways of picking the two numbers one after the other without replacement.
You can pick the prime number in 2 ways and then the multiple of 3 in 3 ways. This gives you 2*3 = 6 ways.
But you can also pick the multiple of 3 first and then the prime number. This gives you another 3*2 = 6 ways.
No of ways of picking the first number is 10 and the second number is 10 again. So total ways = 100 ways
Probability = 12/100
or use probability
20-29 are 10 numbers of which 2 are prime (23 and 29) and 3 are multiples of 3 (21, 24 and 27)
First, you can select a prime with probability 2/10 and then a multiple of 3 with probability 3/10.
Or you can select a multiple of 3 with probability 3/10 and then a prime with probability 2/10.
Total probability = (2/10)*(3/10) + (3/10)*(2/10) = 3/25