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A bag contains 6 tiles numbered from 1 6. If two tiles are selected
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21 Aug 2023, 03:15
21 Aug 2023, 03:15
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Question Stats:
57% (01:37) correct
42% (02:13) wrong based on 7 sessions
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A bag contains 6 tiles numbered from 1 – 6. If two tiles are selected at random, what is the probability that the sum of the numbers on the tiles will be a multiple of 2?
A. 1/5
B. 2/5
C. 3/5
D. 3/4
E. 4/5
Post A Detailed Correct Solution For The Above Questions And Get Kudos.
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A. 1/5
B. 2/5
C. 3/5
D. 3/4
E. 4/5
Post A Detailed Correct Solution For The Above Questions And Get Kudos.
Question From Our Project: Free GRE Prep Club Tests in Exchange for 20 Kudos
\(\Longrightarrow\) GRE - Quant Daily Topic-wise Challenge
\(\Longrightarrow\) GRE Math Essentials - A most comprehensive handout!!
\(\Longrightarrow\) ALL the GRE Quant and Verbal Practice Directories in ONE place (2023)
\(\Longrightarrow\) Best GRE Test Books of 2023
Re: A bag contains 6 tiles numbered from 1 6. If two tiles are selected
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11 Sep 2023, 01:06
11 Sep 2023, 01:06
Is there a faster way for solving the above question without writing down the number of possibilities?
Posted from my mobile device
Posted from my mobile device
A bag contains 6 tiles numbered from 1 6. If two tiles are selected
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11 Sep 2023, 01:25
11 Sep 2023, 01:25
1
Expert Reply
OE
First, we will calculate the total number of ways of selecting two tiles randomly out of 6.
This can be done in \(𝐶^6_2\) ways.
Now, for the sum of numbers to be a multiple of 2, it implies that the sum must be
EVEN.
That means for the sum to be even there are two possibilities:
(i) Either both must be EVEN numbers
(ii) Both must be ODD numbers.
There are 3 EVEN numbered tiles (2, 4, 6) and 3 ODD numbered tiles (1, 3, 5).
Thus, total number of ways of selecting two EVEN number tiles = \(C^3_2\)
Total number of ways of selecting two ODD number tiles =\( C^3_2\)
Therefore,
Total number of ways of selecting 2 even or 2 odd tiles =\(C^3_2+C^3_2\)
Probability will be
\(\dfrac{C^3_2+C^3_2}{C^6_2}=\dfrac{3+3}{\frac{6 \times 5}{2}}=\frac{6}{15}=\frac{2}{5}\)
First, we will calculate the total number of ways of selecting two tiles randomly out of 6.
This can be done in \(𝐶^6_2\) ways.
Now, for the sum of numbers to be a multiple of 2, it implies that the sum must be
EVEN.
That means for the sum to be even there are two possibilities:
(i) Either both must be EVEN numbers
(ii) Both must be ODD numbers.
There are 3 EVEN numbered tiles (2, 4, 6) and 3 ODD numbered tiles (1, 3, 5).
Thus, total number of ways of selecting two EVEN number tiles = \(C^3_2\)
Total number of ways of selecting two ODD number tiles =\( C^3_2\)
Therefore,
Total number of ways of selecting 2 even or 2 odd tiles =\(C^3_2+C^3_2\)
Probability will be
\(\dfrac{C^3_2+C^3_2}{C^6_2}=\dfrac{3+3}{\frac{6 \times 5}{2}}=\frac{6}{15}=\frac{2}{5}\)
