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23 Sep 2020, 02:18
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E
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Difficulty:
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47% (02:17) correct
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If two numbers are selected randomly from the first 100 natural numbers, what is the probability that their sum is divisible by 5?
A. \(\frac{4}{25}\)
B. \(\frac{1}{11}\)
C. \(\frac{1}{20}\)
D. \(\frac{2}{11}\)
E. \(\frac{1}{5}\)
A. \(\frac{4}{25}\)
B. \(\frac{1}{11}\)
C. \(\frac{1}{20}\)
D. \(\frac{2}{11}\)
E. \(\frac{1}{5}\)
23 Sep 2020, 03:22
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[Probability = \frac{Favorable \space Outcomes}{Total \space Outcomes}[/m]
From 1 to 100, we have 10 numbers each ending with 1, 2, 3, 4, 5, 6, 7, 8, 9 and 0
To be divisible by 5, the number should end with a 5 or a 0
A 5 can be obtained as (0 + 5), (1 + 4), (2 + 3), (6 +9) or (7 + 8) = 5 possibilities
Each of these possibilities can have \(^{10}C_1 * ^{10}C_1 = 100\) ways
Total possibilities for numbers ending with 5 = 5 * 100 = 500
A zero can be obtained in 2 ways
(i) (0 + 0) or (5 + 5) = 2 ways, and each of these can be obtained in \(\frac{^{10}C_1 \space * \space ^9C_1}{2}\) = 45 ways
We divide by 2 as order does not matter (to avoid double counting for e.g 5 + 15 is the same as 15 + 5)
Total ways = 2 * 45 = 90
(ii) A zero can be also obtained as the sum of (1 + 9), (2 +8), (3 + 7) or (4 + 6) = 4 ways
Each of these 4 combinations can be obtained in \(^{10}C_1 * ^{10}C_1 = 100\) ways
Total Ways = 4 * 100 = 400
Total possibilities for numbers ending with zero = 90 + 400 = 490
Therefore total numbers divisible by 5 = 500 + 490 = 990
Total Outcomes = \(^{100}C_2 = \frac{100!}{98!2!} = \frac{100*99}{2 * 1} = 4950\)
The Required probability = \(\frac{990}{4950} = \frac{1}{5}\)
Option E
Arun Kumar
From 1 to 100, we have 10 numbers each ending with 1, 2, 3, 4, 5, 6, 7, 8, 9 and 0
To be divisible by 5, the number should end with a 5 or a 0
A 5 can be obtained as (0 + 5), (1 + 4), (2 + 3), (6 +9) or (7 + 8) = 5 possibilities
Each of these possibilities can have \(^{10}C_1 * ^{10}C_1 = 100\) ways
Total possibilities for numbers ending with 5 = 5 * 100 = 500
A zero can be obtained in 2 ways
(i) (0 + 0) or (5 + 5) = 2 ways, and each of these can be obtained in \(\frac{^{10}C_1 \space * \space ^9C_1}{2}\) = 45 ways
We divide by 2 as order does not matter (to avoid double counting for e.g 5 + 15 is the same as 15 + 5)
Total ways = 2 * 45 = 90
(ii) A zero can be also obtained as the sum of (1 + 9), (2 +8), (3 + 7) or (4 + 6) = 4 ways
Each of these 4 combinations can be obtained in \(^{10}C_1 * ^{10}C_1 = 100\) ways
Total Ways = 4 * 100 = 400
Total possibilities for numbers ending with zero = 90 + 400 = 490
Therefore total numbers divisible by 5 = 500 + 490 = 990
Total Outcomes = \(^{100}C_2 = \frac{100!}{98!2!} = \frac{100*99}{2 * 1} = 4950\)
The Required probability = \(\frac{990}{4950} = \frac{1}{5}\)
Option E
Arun Kumar
23 Sep 2020, 03:23
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Explanation:
Total Outcomes = 10x10 = 100
Total favourable outcome= (x,10-x) where x can be 0 too.
so 20 cases (0,10; 1,9, ........... 10,0)
Probability = 20/100 = 1/5
IMO-E
Total Outcomes = 10x10 = 100
Total favourable outcome= (x,10-x) where x can be 0 too.
so 20 cases (0,10; 1,9, ........... 10,0)
Probability = 20/100 = 1/5
IMO-E
