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04 Dec 2022, 09:31
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If five integers are selected from all the integers between 1 and 20, inclusive, without replacement, what is the probability that the five integers will include both 10 and 20?
A. 17/38
B. 1/19
C. 1/20
D. 5/408
E. 1/380
A. 17/38
B. 1/19
C. 1/20
D. 5/408
E. 1/380
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04 Dec 2022, 09:42
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Bunuel
We need 5 integers, and 10 and 20 must be a part of those 5 integers. Hence besides 10 and 20, we need only 3 integers.
Number of ways of selecting 3 integers from 18 integers = 18C3
Note: We are taking 18 integers because 10 and 20 have been taken out.
Total ways of selecting 5 integers from 20 integers = 20C5
Required Probability = \(\frac{18C3 }{ 20C5}\) = \(\frac{1}{19}\)
Option B
26 Jan 2023, 21:34
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Solving using example and theory:
Say the question asks for 1 card only. The probability is 1/20.
Say the question asks to draw 10 cards. Then the probability to get that 1 card is 1/2.
So say the question asks to draw 5 cards. Then the probability to get that 1 card is 5/20, which is 1/4. BUTBUTBUT then, the question also asks you to have another card in the 5-deck; so it'll be (1/19 * 4) ... so together it'll be just 1/19
Say the question asks for 1 card only. The probability is 1/20.
Say the question asks to draw 10 cards. Then the probability to get that 1 card is 1/2.
So say the question asks to draw 5 cards. Then the probability to get that 1 card is 5/20, which is 1/4. BUTBUTBUT then, the question also asks you to have another card in the 5-deck; so it'll be (1/19 * 4) ... so together it'll be just 1/19
