Problem Solving

What is the probability that one of the two integers randomly selected from range 20-29 is prime and the other is a multiple of 3?
(The numbers are selected independently of each other, i.e. they can be equal)



a) 0.06
b) 0.12
c) 0.15
d) 0.18
e) 0.20

[spoiler]
OA is b , but how?? i do not understand a part of the explanation.

Prime integers: 23 and 29. Multiples of 3: 21, 24, and 27.
The probability that the first number is prime while the second is a multiple of 3 = (2/10)*(3/10) = 0.06. [should this not be (2/10)*(3/9) ]
The probability that the first number is a multiple of 3 while the second is prime = (3/10)*(2/10) = 0.06. [should this not be (3/10)*(2/9) ]
The probability that one of the two integers is prime and the other is a multiple of 3 = 0.06 + 0.06 = 0.12 .
[and the resulting sum of probabilities= 2/15 ??][/spoiler]

Prime numbers are 23,29
Multiple of 3 are 21,24,27

Now, probability of picking prime is 2/10
probability of picking mult of 3 is 3/10

probability of both can be done together is 3/10 * 2/10 = 6/100

But the order can change. so 2*6/100 = 12/100 = 0.12

Hope this helps!!

But when i select either of the prime no. or the multiple of three.. wouldn't 9 be left?

so

2/10 * 3/9 and not 2/10 * 3/10

is it that i am not getting the message for the text ??

"The numbers are selected independently of each other, i.e. they can be equal"

arora007 wrote:But when i select either of the prime no. or the multiple of three.. wouldn't 9 be left?

so

2/10 * 3/9 and not 2/10 * 3/10

is it that i am not getting the message for the text ??

"The numbers are selected independently of each other, i.e. they can be equal"

yupp.. numbers are chosen independently.. so you need not change the count.

@arora007,

I agree with your reasoning. The question "must" mention that " the numbers are replaceable" in order to avoid confusion. But your reasoning is correct.

The question is quite poor. Would suggest that you stay off poorly constructed questions such as these.

Enjoy !!!