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16 Oct 2025, 00:28
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A
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Difficulty:
85%
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Question Stats:
54% (03:01) correct
46%
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wrong
based on 131
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For a lottery, three numbers are drawn, each from 1 to 40. A player can win if the product of his or her three numbers is odd. Initially, the same number may be picked up to three times. Later, the rules are changed so that each number may only be picked once. Approximately how much does this reduce a person's probability of winning?
A 0.01
B 0.025
C 0.115
D 0.125
E 0.5
A 0.01
B 0.025
C 0.115
D 0.125
E 0.5
16 Oct 2025, 00:57
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Let's analyze this lottery problem step-by-step:
Numbers are drawn from 1 to 40, with 20 odd and 20 even numbers.
A win requires the product of the three chosen numbers to be odd, so all three numbers must be odd.
Case 1: Repetition allowed
A player picks 3 numbers from 40 with repetition.
Probability = 20*20*20 / 40*40*40 = 0.125.
Case 2: No repetition allowed
A player picks 3 distinct numbers from 40.
Probability = 20*19*18 / 40*39*38 ≈ 0.115
Reduction in probability
0.125−0.115 = 0.01
Hence, A
Numbers are drawn from 1 to 40, with 20 odd and 20 even numbers.
A win requires the product of the three chosen numbers to be odd, so all three numbers must be odd.
Case 1: Repetition allowed
A player picks 3 numbers from 40 with repetition.
Probability = 20*20*20 / 40*40*40 = 0.125.
Case 2: No repetition allowed
A player picks 3 distinct numbers from 40.
Probability = 20*19*18 / 40*39*38 ≈ 0.115
Reduction in probability
0.125−0.115 = 0.01
Hence, A
16 Oct 2025, 06:29
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Bunuel
The lottery numbers are from 1 to 40. The number of odd and even numbers are equal, 20 each.
We need to pick three numbers, such that the product is odd. This means, the number cannot be even.
Case 1: ( Same number can be picked three times).
Picking the first time = 20/40 = 1/2
Second time = 20/40 = 1/2
Third time = 20/40 = 1/2
Probability = (1/2)*(1/2)*(1/2) = 1/8 = 0.125
Case 2: ( Same number cannot be taken)
Probability = 20C3 / (40C3)
= (20*19*18)/(40*39*38)
= 0.115
Difference of probability = 0.125 - 0.115 = 0.01
option A
