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From the even numbers between 1 and 9, two different numbers are to be

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From the even numbers between 1 and 9, two different numbers are to be  [#permalink]

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New post 10 May 2018, 02:17
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Question Stats:

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From the even numbers between 1 and 9, two different numbers are to be chosen at random. What is the probability that their sum will be 8 ?

A. 1/6
B. 3/16
C. 1/4
D. 1/3
E. 1/2

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Re: From the even numbers between 1 and 9, two different numbers are to be  [#permalink]

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New post 10 May 2018, 04:31

Solution



Given:
• From the even numbers between 1 and 9, two different numbers are chosen at random

To find:
• What is the probability that their sum will be 8

Approach and Working:
• Between 1 and 9, the even numbers are 2, 4, 6, and 8
    o The only way sum of two different numbers among them is equal to 8 can be 2 + 6 or 6 + 2

• The first number out of 2 and 6 can be selected in \(\frac{1}{2}\) ways
    o Once the first number is chosen, we need to choose the 2nd number either from (2, 4, 8 – when the 1st chosen number is 6) or from (4, 6, 8 – when then 1st chosen number is 2).

    o Therefore, this choice can be made in \(\frac{1}{3}\) ways

• Hence, the required probability = \(\frac{1}{2} * \frac{1}{3} = \frac{1}{6}\)

Hence, the correct answer is option A.

Answer: A
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Re: From the even numbers between 1 and 9, two different numbers are to be  [#permalink]

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New post 10 May 2018, 05:05
1
Total cases- 9c2 = 36

Favorable cases- (1,7 2,6 3,5) x 2 (for reverse order)

= 6/36

=1/6

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Re: From the even numbers between 1 and 9, two different numbers are to be  [#permalink]

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New post 10 May 2018, 12:08
1
Even numbers between 1 and 9 = 2, 4, 6 & 8.

Total number of ways two numbers can be chosen = 4C2 = 6.

Out of these combinations, only one has the sum = 8 ( 2,6).

Therefore, probability = 1/6. Hence, A.
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From the even numbers between 1 and 9, two different numbers are to be  [#permalink]

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New post 10 May 2018, 18:17
Bunuel wrote:
From the even numbers between 1 and 9, two different numbers are to be chosen at random. What is the probability that their sum will be 8 ?

A. 1/6
B. 3/16
C. 1/4
D. 1/3
E. 1/2


of six total possibilities: 2,4; 2,6; 2,8; 4,6; 4,8; 6,8
only one-2,6-sums to 8
1/6
A
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Re: From the even numbers between 1 and 9, two different numbers are to be  [#permalink]

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New post 13 May 2018, 17:54
Bunuel wrote:
From the even numbers between 1 and 9, two different numbers are to be chosen at random. What is the probability that their sum will be 8 ?

A. 1/6
B. 3/16
C. 1/4
D. 1/3
E. 1/2



The even numbers between 1 and 9 are 2, 4, 6 and 8. So there are 4 even numbers, and we want to choose two of them. The number of ways to do that is 4C2 = (4 x 3)/2 = 6.

The only way to get a sum of 8 from two different numbers is by picking 2 and 6.

We see that the number of favorable outcomes is 1 and the total number of outcomes is 6. Thus, the probability is 1/6.

Answer: A
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Re: From the even numbers between 1 and 9, two different numbers are to be &nbs [#permalink] 13 May 2018, 17:54
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