If p and q are two distinct numbers chosen from the set ..............

Solution

Given:

• A set of 10 integers, {-4, -3, -2, -1, 0, 1, 2, 3, 4, 5}
• p and q are two distinct integers chosen from the given set

To find:

• The probability that p * q is a prime number

Approach and Working:
The total number of ways of choosing two distinct integers from a set of 10 elements = \(^{10}C_2 = 45\)

Now, for p * q to be prime, one of them must be ±1 and the other must be = ± a prime number.

• Thus, the possible cases are {-1, -2}, {-1, -3}, {1, 2}, {1, 3}, {1, 5}

Therefore, probability = \(\frac{5}{45} = \frac{1}{9}\)

Hence the correct answer is Option C.

Answer: C


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prime factors in the set: -3, -2, 2, 3, 5
Each product with 1 or -1 is also prime.
5*2/10C2=10/45=2/9→(E)

Posted from my mobile device

prime no are +ve integer value (2,3,5)
2,3,5 can be formed ( -3,-1) ,( -2,-1), ( 1,2), ( 1,3),(1,5) ; total 5 pairs p*q can be 2 ways; 5*2 = 10

total pairs = 5 and p*q = 10c2
5/10c2 ; 5/45
=1/9
IMOC

I still dont understand why this probability approach doesnt work:-
p(getting 1)xp(getting 2/3/5) + p(getting -1)xp(getting -2, -3)
= 1/10 x 3/9 + 1/10 x 2/9
=3/90 + 2/90
= 5/90
=1/18
can anyone explain? is it because it needs to be multiplied with 2 ie. (1/18)x2 because order of picking the the numbers matters?

chetan2u, Bunuel, VeritasKarishma

I always get caught in questions such as these-

For instance: if we have to select 2 numbers to obtain an odd no, the first number can be even and second odd or first odd and second even. I applied the same principle and got 2/9. What am I missing here??

If you need an odd product, both numbers should be odd only.

Probability = 5C2 / 9C2

or

Probability = (5/9)*(4/8)
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VeritasKarishma

Sorry for not explaining my question clearly.

the possible cases are {-1, -2}, {-1, -3}, {1, 2}, {1, 3}, {1, 5}. Should we not include {-2, -1}, {-3, -1}, {2, 1}, {3, 1}, {5, 1}

Thus the total cases should be 10. But the solution posted has not taken into account all the cases. What is wrong with analysis?

Reference your PM.
You can take any but ensure that TOTAL ways is also in same C or P.
Here you are taking Permutations as -1,-3 and -3,-1 are different.
So the total ways will also be 10*9.. first pick any of 10 and next any of remaining 9.
Probability =10/(10*9)=1/9
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This is a probability question. If you arrange in the numerator, you need to arrange in the denominator too. If you do not arrange, you do not arrange in either.

If I say the two numbers are distinct - p and q - then I can pick them both in 10 ways such that pq is prime (as you have done (-1, -2), (-1, -3) ... etc)
But then, p can be picked in 10 ways and q can be picked up in 9 ways leading to 90 overall ways of picking the two numbers.

Then Probability = 10/90 = 1/9

On the other hand, if we do as done above - 5 ways of picking the two numbers without arranging so (-1, -2) is the same as (-2, -1) and will be counted once only, we pick the 2 numbers in 10C2 = 45 ways (no arrangement of p and q here)

Then probability = 5/45 = 1/9
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there are 10 possible sets for ordered pair of (p,q) which are {-1, -2}, {-1, -3}, {1, 2}, {1, 3}, {1, 5}, {-2, -1}, {-3, -1}, {2, 1}, {3, 1}, {5, 1}
Here, the order matters, so it is permutation

Hence, total possible outcomes = 10P2 = 90

required probability = 10/90 = 1/9

C is correct.

Should 0 be excluded from the denominator?

Prime nos are positive integers. They can't be negative.

While you arrived at the right solution, I am not sure how you arrived at the conclusion that this is a case of Permutation? Could you please elaborate.

As per my understanding, we are "selecting" two nos from the given set. And since we are interested in the product of those two nos (to be Prime), the two sets
(-2,-1) and (-1, -2) give the same result. Therefore, the order in which the two nos are chosen does not matter.

Let me know if I have missed out on something here?

I will answer this in two parts
1st part
Lets understand my reasoning
We know Probability = Desired number of cases / total number of cases

Now, in desired number of cases ie 10, I have considered ordered pair ie. order matter, in other words (-1,-2) is different from (-2,-1). please refer to the set above.

Similarly, to calculate total number of cases, I would have to consider permutation ie order matters (and not combination), therefore 10P2

Required prob = 1/9

2nd part
If you still have to find out the answer with combination, that also is possible, lets look into it

Now desired number of cases will 5 ie. {-1, -2}, {-1, -3}, {1, 2}, {1, 3}, {1, 5}. pl see I have not taken (-1,-2) different from (-2,-1), in other words order does not matter here.

Total number of cases can be 10C2 = 45

Required prob = 5/45 = 1/9 which is same

I hope now it is clear!

kapil1

Yes, got it. Thanks!

Why should we count the product p*q=0, 18 times, and not only 1 time?
Doesn't it duplicate the outcomes?

Hi Karishma,
I want to know what is wrong in my understanding of the total outcomes in the denominator. As we need to find probability of p*q being a prime number, my sense is the denominator should be all unique values of the product p*q out of which probability of prime number is to be found. By unique values I mean, p*q can be 0 if either of one is 0, if either is 1 or -1, the values of p*q will be duplicate or some not for eg: p=1& q=2 or p=-1 & q=-2.
When we consider 10C2, we are not eliminating the duplicates for eg: p*q = 0 when 0 is multiplied with other unique number from the set. I hope you got my query.

Hi ,
I want to know what is wrong in my understanding of the total outcomes in the denominator. As we need to find probability of p*q being a prime number, my sense is the denominator should be all unique values of the product p*q out of which probability of prime number is to be found. By unique values I mean, p*q can be 0 if either of one is 0, if either is 1 or -1, the values of p*q will be duplicate or some not for eg: p=1& q=2 or p=-1 & q=-2.
When we consider 10C2, we are not eliminating the duplicates for eg: p*q = 0 when 0 is multiplied with other unique number from the set. I hope you got my query.

Hi e-gmat quant expert,
I want to know what is wrong in my understanding of the total outcomes in the denominator. As we need to find probability of p*q being a prime number, my sense is the denominator should be all unique values of the product p*q out of which probability of prime number is to be found. By unique values I mean, p*q can be 0 if either of one is 0, if either is 1 or -1, the values of p*q will be duplicate or some not for eg: p=1& q=2 or p=-1 & q=-2.
When we consider 10C2, we are not eliminating the duplicates for eg: p*q = 0 when 0 is multiplied with other unique number from the set. I hope you got my query.