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Zorluk: ZorComplementary Probability and At-Least-One Scenarios

If 3 components are randomly selected without replacement from a batch of 10 components containing exactly 2 defective components, the probability that at least one selected component is defective is equal to 815\frac{8}{15}.

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True. The probability of selecting at least one defective component is indeed 815\frac{8}{15}.
The statement correctly computes the probability of selecting at least one defective component using 1−P(no defective components)=1−(83)(103)=1−715=8151 - P(\text{no defective components}) = 1 - \frac{\binom{8}{3}}{\binom{10}{3}} = 1 - \frac{7}{15} = \frac{8}{15}.

Adım Adım Çözüm

1
Identify the complementary event
The complement of 'at least one defective component' is 'zero defective components' (all 3 selected components are non-defective).
Calculating P(at least one)=1−P(none)P(\text{at least one}) = 1 - P(\text{none}) avoids calculating multiple dependent individual outcomes.
2
Calculate the total combinations for choosing 3 components from 10
(103)=10×9×83×2×1=120\binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120.
This establishes the sample space size.
3
Calculate the combinations for choosing 3 non-defective components from 8
(83)=8×7×63×2×1=56\binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56.
There are 8 non-defective components in the batch.
4
Compute P(0 defective)P(\text{0 defective}) and subtract from 1
P(0 defective)=56120=715P(\text{0 defective}) = \frac{56}{120} = \frac{7}{15}. Therefore, P(at least 1 defective)=1−715=815P(\text{at least 1 defective}) = 1 - \frac{7}{15} = \frac{8}{15}.
Completing the complementary calculation confirms the statement is correct.

Anahtar Kavram

Complementary Probability and At-Least-One Scenarios without Replacement
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Complementary Probability and At-Least-One Scenarios

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