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Difficulty:
95%
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Question Stats:
39% (03:08) correct
61%
(02:52)
wrong
based on 99
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Tina randomly selects 2 distinct numbers from the set {1, 2, 3, 4, 5} and Sergio randomly selects a number from the set {1, 2, . . ., 10}. What is the probability that the number Sergio chooses is larger than the sum of the two numbers Tina chooses?
A. \(\frac{2}{5}\)
B. \(\frac{9}{20}\)
C. \(\frac{1}{2}\)
D. \(\frac{11}{20}\)
E. \(\frac{24}{25}\)
A. \(\frac{2}{5}\)
B. \(\frac{9}{20}\)
C. \(\frac{1}{2}\)
D. \(\frac{11}{20}\)
E. \(\frac{24}{25}\)
17 Feb 2022, 02:39
Kudos
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Let me try to solve the question with a hard, but straightforward way:
1. If Tina chooses 2 distinct numbers from 5 numbers, then there will be \(\frac{5!}{2!3!}\) = 10 ways. Their sum are:
1 + 2 = 3
1 + 3 = 4
1 + 4 = 5
1 + 5 = 6
2 + 3 = 5
2 + 4 = 6
2 + 5 = 7
3 + 4 = 7
3 + 5 = 8
4 + 5 = 9
We can simply say that Tina randomly chooses a number from these 10 numbers.
Then Sergio chooses a number from the list {1, 2, ... 10). He also has 10 choices. We can calculate the probability of Sergio's number being larger than Tina's number for each choice of Tina and average them, since they all have equal chances. For example, if Tina chooses 3, then there is \(\frac{7}{10}\) probability that Sergio's number will be larger than 3 since he has 7 numbers larger than 3 (4-10). In the same way we calculate probability for each choice of Tina:\((\frac{3}{10} + \frac{4}{10} + \frac{5}{10} + \frac{6}{10} + \frac{5}{10} + \frac{6}{10} + \frac{7}{10} + \frac{7}{10} + \frac{8}{10} + \frac{9}{10}) * \frac{1}{10}\) = \(\frac{(3 + 4 + 5 + 6 + 5 + 6 + 7 + 7 + 8 + 9)}{100}\) = \(\frac{40}{100} = \frac{2}{5} \) A)
1. If Tina chooses 2 distinct numbers from 5 numbers, then there will be \(\frac{5!}{2!3!}\) = 10 ways. Their sum are:
1 + 2 = 3
1 + 3 = 4
1 + 4 = 5
1 + 5 = 6
2 + 3 = 5
2 + 4 = 6
2 + 5 = 7
3 + 4 = 7
3 + 5 = 8
4 + 5 = 9
We can simply say that Tina randomly chooses a number from these 10 numbers.
Then Sergio chooses a number from the list {1, 2, ... 10). He also has 10 choices. We can calculate the probability of Sergio's number being larger than Tina's number for each choice of Tina and average them, since they all have equal chances. For example, if Tina chooses 3, then there is \(\frac{7}{10}\) probability that Sergio's number will be larger than 3 since he has 7 numbers larger than 3 (4-10). In the same way we calculate probability for each choice of Tina:\((\frac{3}{10} + \frac{4}{10} + \frac{5}{10} + \frac{6}{10} + \frac{5}{10} + \frac{6}{10} + \frac{7}{10} + \frac{7}{10} + \frac{8}{10} + \frac{9}{10}) * \frac{1}{10}\) = \(\frac{(3 + 4 + 5 + 6 + 5 + 6 + 7 + 7 + 8 + 9)}{100}\) = \(\frac{40}{100} = \frac{2}{5} \) A)
01 Sep 2023, 05:46
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twobagels
Tina can choose 2 distinct numbers from the set {1, 2, 3, 4, 5} in 5c2 = 10 ways
Sergio randomly selects a number from the set {1, 2, . . ., 10} in 10 ways
Overall, they can make choices in 100 ways.
the number Sergio chooses is larger or smaller or equal to the sum of the two numbers Tina chooses
Sum of Tina numbers lie in [3,4,5,6,5,6,7,7,8,9]
So, probability of the numbers being equal + larger + smaller = 1
Sergio chooses a number equal to or smaller than [3,4,5,6,5,6,7,7,8,9] in [3,4,5,6,5,6,7,7,8,9] ways respectively.
For example, he can choose a number less than or equal to 3 in 3 ways by choosing 1,2 or 3.
Totally, in sum([3,4,5,6,5,6,7,7,8,9]) ways = 60 ways
Sergio chooses a number greater than [3,4,5,6,5,6,7,7,8,9] in 100-60 cases = 40
Probability = 40/100 = 2/5
