Let A be the vent that a randomly selected two digit number

Question

Let A be the event that a randomly selected two digit number is divisible by 3 and let B be the event that a randomly selected two digit number is divisible by 5. What is P(A and B)?

A. 1/18
B. 1/15
C. 1/5
D. 1/3
E. 1/2

ENGRTOMBA2018 · Answer

Please format the question properly and mention anything related to the solution either in another post or put it under "spoiler" as shown above.

For your question,

For 2 independent events ,

P(A and B) = P(A)*P(B)

where 
P(A) = selecting a multiple of 3 in all the 2 digit numbers = Favorable cases/ Total cases = 30/90 = 1/3
P(B) = selecting a multiple of 5 in all the 2 digit numbers = Favorable cases/ Total cases = 18/90 = 1/5

Thus, P(A and B) = P(A)*P(B) = (1/3)(1/5) = 1/15 . B is the correct answer.

ashokk138 · Answer

We know that

P(A and B) = P(A) * P(B)

Event where a two digit no is divisible by 3 => 30 (12 to 99)
Total two digit no.s = 90
Hence P(A) = 30/90 => 1/3

Similarly, Event where a two digit no is divisible by 5 => 18 (10 to 95)
P(B) = 18/90 => 1/5

P(A and B) = 1/3 * 1/5 => 1/15

Option B

appsy01 · Answer

Given that P(A) be the probability that a number divisible by 3 and P(B) be the probability that a number is divisible by 5

=> P ( A and B ) is the number divisible by both 3 and 5

the Numbers are 15, 30, 45, 60, 75, 90 => 6

Total No of Events = Count (10 to 99) = 90

Therefore  P(A and B) = \frac{Favorable Events}{Total Events}  = \frac{6}{90} = \frac{1}{15}

Ans: B

MathRevolution · Answer

Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.

Let A be the event that a randomly selected two digit number is divisible by 3 and let B be the event that a randomly selected two digit number is divisible by 5. What is P(A and B)?

A. 1/18
B. 1/15
C. 1/5
D. 1/3
E. 1/2

The event A and B means that a randomly selected two digit number is divisible by both 3 and 5. Since the least common multiple of 3 and 5 is 15, that event is same as a randomly selected two digit number is divisible by 15.

The total number of two digit number is 90(from 10 to 99 inclusively). The number of the multiples of 15 from 10 to 99 is  =6.

So P(A and B) = 6/90 = 3/45 = 1/15. So the answer is B.

mvictor · Answer

can't believe I missed this one because of a silly mistake...
we have 90 numbers.
we need positive outcomes when the numbers are divisible by 3 and 5.
we have: 15, 30, 45, 60, 75, 90 - 6 numbers.
6/90 = 1/15

Archit3110 · Answer

total 2 digits ; 90
and P its divisible by 3 ; 30/90 ; 1/3
and P its divisible by 5 ; 18/90 ; 1/5
P(A and B) both 3 & 5 divisible 1/3 *1/5 ; 1/15
OPTION B

Gauravji21 · Answer

There are two ways of doing it 
 Way 1 :  Finding P(a) & P(b) and then multiplying both probabilities to get P(a & b)
P(A) = two digit numbers divisible by 3 / total number of 02 digit numbers = 30 / 90 = 1/3 ( here you don't need to write and count the number of two digit numbers divisible by 3. Keep in mind that every third consecutive number is divisible by 3. So starting from 10 to 99 we have a total of 90 digits. one third will be divisible by 3 i.e. 90/3 = 30)
P(B) = two digit numbers divisible by 5 / total number of 02 digit numbers = 18 / 90 = 1/5 (Literally we need to write the multiple of 5 which are of two digits and count them)

Way 2 :  Directly finding P(a&b) i.e the probability of randomly selecting two digit number which is divisible by both 3 & 5.
Such numbers are multiples of LCM of 3 & 5 i.e. 15, 30, 45, 60, 75 , 90. So, in total 06 numbers
Total number of 02 digit numbers = 90 ( from 100 you can remove single digit numbers and three digit number-1 to 9 & 100)
So, P(A&B) = 6/90 = 2 / 30 = 1 /15

IanStewart · Answer

Roughly half of the solutions in this thread are not correct (the ones that are multiplying two probabilities). The "rule"

P(A and B) = P(A) * P(B)

is only true when A and B are independent. So you can use that rule if you're rolling a die and then flipping a coin, say, because then the outcome of the first event has no influence on the outcome of the second. But if you're concerned with just a single event, like picking a single number from a set, you're absolutely never dealing with two independent events, so you can never use this rule (or at least not without risking a mistake).

You can see why this multiplication approach is not correct just by imagining a different set - say we pick one number from this list:

1, 2, 3, 4, 5, 6, 7, 8, 9, 10

and want to know if we pick a number that is both a multiple of 3 and a multiple of 5. Picking one number randomly from this set, the probability of picking a multiple of 3 is 3/10, and of picking a multiple of 5 is 1/5, but the probability of picking a number divisible by both 3 and 5 (so by 15) is not (3/10)(1/5) = 3/50. It is zero.

The solutions in this thread that are correct are those that are counting the number of multiples of 15 we have (there are six of them), and then dividing by the size of the set we're picking from (there are 90 numbers we're selecting from).

Gauravji21 · Answer

Thanks  IanStewart  for correcting me. I had suggested two ways above. Now, one of them is incorrect.
I was lucky in this case as I got same answer both ways. Else I would have marked wrong answer.

amritbharadwaj · Answer

Events A and B can be considered approximately independent. With this assumption:
P(A): The probability of a two-digit number being divisible by 3 is 1/3 (30/90).
P(B): The probability of a two-digit number being divisible by 5 is 1/5 (18/90).

Therefore, assuming independence:
P(A and B) = P(A) * P(B) = 1/3 * 1/5 = 1/15
This means that there is a 1/15 probability that a randomly selected two-digit number is both divisible by 3 and 5.

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Let A be the vent that a randomly selected two digit number

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