What is the probability that rs=r?

Hi Bunnel ,

Can you please clarify one thing ,we are breaking the probability (1/9) in the form 1/3 and 1/3 becoz we have 3 elements in the set :-

Or else if the no of elements wasn't given then we could have broken the same in various other ways like 1/6 *2/3 or other ,then b would also have been insufficient.

Is my understanding correct ?

Thanks in advance.

Countdown

Notice that not only we are told that the sets contain 3 integers each but we are also told that each set contains three distinct positive integers. From (2) we have that R={1, a, b} and S={1, c, d}, where neither from a, b, c, and d is 1. So, P(1 from R)=P(1 from S)=1/3.

Hope it's clear.

I know i am wrong but kindly do explain why cant i read the statement above as R and S both contain Three distinct positive integers.

R = {a, b , c}
S = {p, q, r}

Is it because its not explicitly mentioned that there is no overlap?

Sorry, but I don't understand your questions at all...

Sorry i had a mistake in my last post...Why cant i Assume that

R = {r, a, b}
S = {s, c, d}

Where a,b,c,d,r,s are all distinct?

If that is the case then

We have to check if s= 1

A - s(r-1) = 0 s cant be zero so r is 1.. If r is 1, s cant be one as all are distinct - Sufficient

B - r + s = 2 1/9 Impossible hence insufficient?

We are told that sets R and S each contain three distinct positive integers. This does not mean that all 6 integers are distinct.

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Sets R and S each contain three distinct positive integers. If integer r is randomly selected from R and integer s is randomly selected from S, what is the probability that rs = r?

(1) The probability that rs = s is 1/3

(2) The probability that r + s = 2 is 1/9

We can modify the question into: rs=r, rs-r=0. r(s-1)=0,and ultimately whether r=0 or s=1? But the question is regarding distinct positive integers, so what we really want to know is whether s=1.
For condition 2, the only situation where r+s=2 is when r=s=1, so the probability that s=1 is only 1/9, making the condition unique and sufficient,
Whereas condition 1 gives s(r-1)=0 and the probability that r=1 is 1/3, but we cannot know the probability that s=1, so this condition is insufficient, making the answer (B).

Once we modify the original condition and the question according to the variable approach method 1, we can solve approximately 30% of DS questions.

Both P and Q each contain 3 distinct positive integers. if rs=r , s has to be equal to 1.
St 1 - NS , we don't know if s=1
St 2 - Sufficient ; r=s=1. Elements is s are distinct. Hence we can choose any element of r but we should choose only one element of s. The probability in such a case is 1/3.

Option B it is !

The way I interpreted the problem:

We are given that each of the 3 integers in each set are distinct and POSITIVE Integers.

The questions asks us: “what is the probability that: RS = R?”

Or

RS - R = 0

R (S - 1) = 0

Zero Product Rule: either R = 0 or S = 1

Since we have only positive integers, the only way for this to be possible is if S = 1.

Since all 3 integers WITHIN a Set must be distinct, there are 2 possibilities:

Either: (case 1) +1 is one of the three Distinct positive integers in Set S

Or: (case 2) +1 does NOT exist in Set S

It does not matter which of the 3 positive integers we pick from Set R. All that matters, in order to get the favorable outcome, is that we pick the +1 from Set S.

Case 1: Set S {1 , A , B} —— where A and B are distinct and do not equal 1

Probability that (RS = R) = (1/3)

Or

Case 2: Set S {X , Y , Z} ——- where X, Y. Z are distinct positive integers and NONE of them is equal to +1

Probability that (RS = R) = 0 (no chance)

In other words, the contents of Set R are irrelevant. We want to know whether Set S contains +1 as one of its three distinct positive integers.

S1: tells us that Set R must contain +1

For RS = S, given that these are all positive integers —— R must = +1

So set R contains {1 , P, Q} —- where P and Q are distinct positive integers that do not equal +1

We can select any number from Set S (3/3)
And
Probability of selecting the 1 from set R is = (1/3)

Hence, probability that RS = S is (1/3)

Does not tells us whether Set S does or does NOT contain +1

Not sufficient.

Statement 2:

For R + S = 2 ——- given that R and S each must be Positive Integers, there is only one possibility:

R = 1 —— and —— S = 1

Since the probability that R + S = 2 is not zero, we know that Set S MUST contain +1 as one its three distinct positive integers.

And the probability of pulling a +1 from set S is therefore = (1/3)

Which means the probability that RS = R is also (1/3)

Statement 2 is sufficient alone.
B



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Hi Bunuel,
for option B you said P(1&1) = P(1 from R)*P(1 from S)=1/3*1/3=1/9

but shouldn't it be ---> P(1&1) = 2*1/3*1/3 ( as you take 1st R then S OR 1st S and then R)

The selection of 1 from set R and 1 from set S are independent events and the order of selection doesn't affect the outcome.

The multiplication by 2 that you suggested is typically used in problems where the order of events matters or when there are two distinct ways to achieve the same outcome. In this case, whether you pick 1 from R first and then S, or vice versa, doesn't change the overall event which is picking 1 from both R and S.

To illustrate this with an example, imagine we have a box containing 1 red, 1 blue, 1 black, and 1 white ball. If we were to calculate the probability of drawing 1 red and 1 blue ball without replacement, the formula would be 2*(1/4)*(1/3). This takes into account the two distinct sequences in which these selections could occur: red then blue, or blue then red.

In contrast, if we had two separate boxes - one holding 1 red and 1 black ball, and the other containing 1 blue and 1 white ball - the probability of drawing a red ball from the first box and a blue ball from the second box is simply (1/2)*(1/2). This calculation does not require a multiplication by 2, as the sequence of selections between the two boxes does not affect the outcome.

Hope it's clear.