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Manager  G
Joined: 26 Dec 2015
Posts: 248
Location: United States (CA)
Concentration: Finance, Strategy
WE: Investment Banking (Venture Capital)
If x is chosen randomly from the set {7, 8, 9, 10, 11} and y is chosen  [#permalink]

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3 00:00

Difficulty:   5% (low)

Question Stats: 84% (00:49) correct 16% (00:51) wrong based on 113 sessions

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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) $$\frac{1}{10}$$
Manager  G
Joined: 22 May 2015
Posts: 109
Re: If x is chosen randomly from the set {7, 8, 9, 10, 11} and y is chosen  [#permalink]

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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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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
Manager  G
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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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2
1
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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Manager  S
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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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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

Give kudos if you like the answer Re: If x is chosen randomly from the set {7, 8, 9, 10, 11} and y is chosen   [#permalink] 28 Dec 2018, 21:40
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