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If x is chosen randomly from the set {7, 8, 9, 10, 11} and y is chosen

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If x is chosen randomly from the set {7, 8, 9, 10, 11} and y is chosen  [#permalink]

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New post 18 Aug 2017, 08:24
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If x is chosen randomly from the set {7, 8, 9, 10, 11} and y is chosen randomly from the set {20, 21, 22, 23}, what is the probability that both x and y are prime numbers?

A) \(\frac{1}{5}\)

B) \(\frac{3}{20}\)

C) \(\frac{13}{20}\)

D) \(\frac{3}{10}\)

E) \(\frac{1}{10}\)
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Re: If x is chosen randomly from the set {7, 8, 9, 10, 11} and y is chosen  [#permalink]

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New post 18 Aug 2017, 08:31
1
Gng back to basics

Probability = No of Favorable Outcomes / total no of Outcomes

Total No of Outcomes = 5C1 * 4C1
No of Favorable Outcomes = 2 * 1

=> Probability = 1/10
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Re: If x is chosen randomly from the set {7, 8, 9, 10, 11} and y is chosen  [#permalink]

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New post 18 Aug 2017, 11:15
LakerFan24 wrote:
If x is chosen randomly from the set {7, 8, 9, 10, 11} and y is chosen randomly from the set {20, 21, 22, 23}, what is the probability that both x and y are prime numbers?

A) \(\frac{1}{5}\)

B) \(\frac{3}{20}\)

C) \(\frac{13}{20}\)

D) \(\frac{3}{10}\)

E) \(\frac{1}{10}\)


For \(x\) there are 2 prime nos available in the set of 5 integers {\(7,11\)}. Hence Probability of \(x\) being prime \(= \frac{2}{5}\)
For \(y\) there is 1 prime no. available in the set of 4 integers {\(23\)}. Hence Probability of \(y\) being prime \(= \frac{1}{4}\)
So Probability of \(x AND y\) being prime \(= \frac{2}{5}*\frac{1}{4} = \frac{1}{10}\)
Option E
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If x is chosen randomly from the set {7, 8, 9, 10, 11} and y is chosen  [#permalink]

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New post 19 Aug 2017, 20:54
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LakerFan24 wrote:
If x is chosen randomly from the set {7, 8, 9, 10, 11} and y is chosen randomly from the set {20, 21, 22, 23}, what is the probability that both x and y are prime numbers?

A) \(\frac{1}{5}\)

B) \(\frac{3}{20}\)

C) \(\frac{13}{20}\)

D) \(\frac{3}{10}\)

E) \(\frac{1}{10}\)


This is AND question - so we must MULTIPLY each probability.
1. P (X is prime) = \(\frac{possibilites that X are prime}{total number in the set}\) = \(\frac{2}{5}\). Only 7 and 11 are prime.
2. P (Y is prime) = \(\frac{possibilites that Y are prime}{total number in the set}\) = \(\frac{1}{4}\). Only 23 is prime.
3. P (X is prime AND y is prime) = \(\frac{2}{5}\) * \(\frac{1}{4}\) = \(\frac{2}{20}\)

E.
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Re: If x is chosen randomly from the set {7, 8, 9, 10, 11} and y is chosen  [#permalink]

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New post 28 Dec 2018, 20:40
septwibowo wrote:
LakerFan24 wrote:
If x is chosen randomly from the set {7, 8, 9, 10, 11} and y is chosen randomly from the set {20, 21, 22, 23}, what is the probability that both x and y are prime numbers?

A) \(\frac{1}{5}\)

B) \(\frac{3}{20}\)

C) \(\frac{13}{20}\)

D) \(\frac{3}{10}\)

E) \(\frac{1}{10}\)


This is AND question - so we must MULTIPLY each probability.
1. P (X is prime) = \(\frac{possibilites that X are prime}{total number in the set}\) = \(\frac{2}{5}\). Only 7 and 11 are prime.
2. P (Y is prime) = \(\frac{possibilites that Y are prime}{total number in the set}\) = \(\frac{1}{4}\). Only 23 is prime.
3. P (X is prime AND y is prime) = \(\frac{2}{5}\) * \(\frac{1}{4}\) = \(\frac{2}{20}\)

E.


Ways to select prime number 7 and 11 from the first set is 2/5
Ways to sleect prime number 23 from second set 1/4
combine probability is P(A)*P(B) = (2/5)*(1*4) = 1/10


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Re: If x is chosen randomly from the set {7, 8, 9, 10, 11} and y is chosen &nbs [#permalink] 28 Dec 2018, 20:40
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