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

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If p and q are two distinct numbers chosen from the set ..............  [#permalink]

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New post 30 Jan 2019, 00:43
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If p and q are two distinct numbers chosen from the set {-4,-3, -2, -1, 0, 1, 2, 3, 4, 5}. Find the probability that p * q is a prime number.

    A. \(\frac{1}{30}\)

    B. \(\frac{1}{15}\)

    C. \(\frac{1}{9}\)

    D. \(\frac{2}{15}\)

    E. \(\frac{2}{9}\)


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Re: If p and q are two distinct numbers chosen from the set ..............  [#permalink]

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New post 30 Jan 2019, 01:07
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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)

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If p and q are two distinct numbers chosen from the set ..............  [#permalink]

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New post Updated on: 01 Feb 2019, 08:21
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
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EgmatQuantExpert wrote:
If p and q are two distinct numbers chosen from the set {-4,-3, -2, -1, 0, 1, 2, 3, 4, 5}. Find the probability that p * q is a prime number.

    A. \(\frac{1}{30}\)

    B. \(\frac{1}{15}\)

    C. \(\frac{1}{9}\)

    D. \(\frac{2}{15}\)

    E. \(\frac{2}{9}\)


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Originally posted by Archit3110 on 30 Jan 2019, 06:04.
Last edited by Archit3110 on 01 Feb 2019, 08:21, edited 1 time in total.
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Re: If p and q are two distinct numbers chosen from the set ..............  [#permalink]

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New post 01 Feb 2019, 00:57
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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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Re: If p and q are two distinct numbers chosen from the set ..............  [#permalink]

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New post 26 Oct 2019, 01:20
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?
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New post 31 Dec 2019, 22:40
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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??
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Re: If p and q are two distinct numbers chosen from the set ..............  [#permalink]

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New post 02 Jan 2020, 03:49
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delta23 wrote:
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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New post 04 Jan 2020, 22:22
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?
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Re: If p and q are two distinct numbers chosen from the set ..............  [#permalink]

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New post 05 Jan 2020, 00:31
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delta23 wrote:
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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Re: If p and q are two distinct numbers chosen from the set ..............  [#permalink]

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New post 05 Jan 2020, 21:58
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delta23 wrote:
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?


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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If p and q are two distinct numbers chosen from the set ..............  [#permalink]

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New post 05 Jan 2020, 23:44
EgmatQuantExpert wrote:
If p and q are two distinct numbers chosen from the set {-4,-3, -2, -1, 0, 1, 2, 3, 4, 5}. Find the probability that p * q is a prime number.

    A. \(\frac{1}{30}\)

    B. \(\frac{1}{15}\)

    C. \(\frac{1}{9}\)

    D. \(\frac{2}{15}\)

    E. \(\frac{2}{9}\)


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.
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If p and q are two distinct numbers chosen from the set ..............   [#permalink] 05 Jan 2020, 23:44
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