We have a batch of 100 musical instruments, 4 of which have impure tone. We randomly select 3 instruments for testing. If at least one of the 3 is identified as having impure tone, the whole batch is rejected. The probability of correctly identifying an impure instrument is 0.95. The probability of mistakenly identifying a pure instrument as impure is 0.01. We want to find the probability that the batch is accepted.

1. Problem Description

We have a batch of 100 musical instruments, 4 of which have impure tone. We randomly select 3 instruments for testing. If at least one of the 3 is identified as having impure tone, the whole batch is rejected. The probability of correctly identifying an impure instrument is 0.

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5. The probability of mistakenly identifying a pure instrument as impure is 0.

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1. We want to find the probability that the batch is accepted.

2. Solution Steps

The batch is accepted if all 3 tested instruments are identified as having pure tone. This can happen in two ways: (1) All 3 are actually pure and are identified as pure; or (2) at least one is impure, but all are identified as pure.

Let $I$ denote an impure instrument and $P$ denote a pure instrument.

Let $I'$ denote that an instrument is identified as impure, and $P'$ denote that an instrument is identified as pure.

The probability of selecting a pure instrument is $\frac{100-4}{100} = \frac{96}{100} = 0.96$.

The probability of selecting an impure instrument is $\frac{4}{100} = 0.04$.

$P(P'|P) = 1 - 0.01 = 0.99$. This is the probability that a pure instrument is identified as pure.

$P(P'|I) = 1 - 0.95 = 0.05$. This is the probability that an impure instrument is identified as pure.

For the batch to be accepted, we need all 3 instruments tested to be identified as pure. Let A be the event that the batch is accepted.

There are $\binom{100}{3}$ ways to choose 3 instruments.

The number of ways to select 3 pure instruments is $\binom{96}{3} = \frac{96 \cdot 95 \cdot 94}{3 \cdot 2 \cdot 1} = 142880$.

The number of ways to select 2 pure instruments and 1 impure instrument is $\binom{96}{2} \binom{4}{1} = \frac{96 \cdot 95}{2} \cdot 4 = 18240$.

The number of ways to select 1 pure instrument and 2 impure instruments is $\binom{96}{1} \binom{4}{2} = 96 \cdot \frac{4 \cdot 3}{2} = 96 \cdot 6 = 576$.

The number of ways to select 3 impure instruments is $\binom{4}{3} = 4$.

Total ways to select 3 instruments is $\binom{100}{3} = \frac{100 \cdot 99 \cdot 98}{3 \cdot 2 \cdot 1} = 161700$.

Note that $142880 + 18240 + 576 + 4 = 161700$, as expected.

$P(\text{all 3 pure}) = \frac{\binom{96}{3}}{\binom{100}{3}} = \frac{142880}{161700} \approx 0.8836$. The probability that all 3 are pure and all identified as pure is $0.8836 \cdot (0.99)^3 \approx 0.8576$

$P(\text{2 pure, 1 impure}) = \frac{\binom{96}{2} \binom{4}{1}}{\binom{100}{3}} = \frac{18240}{161700} \approx 0.1128$. The probability that 2 are pure, 1 is impure, and all are identified as pure is $0.1128 \cdot (0.99)^2 \cdot (0.05) \approx 0.0055$.

$P(\text{1 pure, 2 impure}) = \frac{\binom{96}{1} \binom{4}{2}}{\binom{100}{3}} = \frac{576}{161700} \approx 0.0036$. The probability that 1 is pure, 2 are impure, and all are identified as pure is $0.0036 \cdot (0.99) \cdot (0.05)^2 \approx 0.000009$.

$P(\text{all 3 impure}) = \frac{\binom{4}{3}}{\binom{100}{3}} = \frac{4}{161700} \approx 0.000025$. The probability that all 3 are impure and all are identified as pure is $0.000025 \cdot (0.05)^3 = 0.000025 \cdot 0.000125 \approx 0.0000000031$.

The probability the batch is accepted is approximately $0.8576 + 0.0055 + 0.000009 + 0.0000000031 \approx 0.8631$.

Alternative solution:

We want to find the probability that all 3 selected instruments are identified as having pure tone. This is the same as the probability that we do not reject the batch.

$P(\text{accept}) = P(\text{all 3 identified as pure})$.

$P(\text{all 3 identified as pure}) = \sum_{k=0}^{3} P(\text{all 3 identified as pure} | k \text{ impure}) P(k \text{ impure})$, where $k$ is the number of impure instruments chosen.

$P(0 \text{ impure}) = \frac{\binom{96}{3}}{\binom{100}{3}} = \frac{142880}{161700}$.

$P(1 \text{ impure}) = \frac{\binom{4}{1} \binom{96}{2}}{\binom{100}{3}} = \frac{18240}{161700}$.

$P(2 \text{ impure}) = \frac{\binom{4}{2} \binom{96}{1}}{\binom{100}{3}} = \frac{576}{161700}$.

$P(3 \text{ impure}) = \frac{\binom{4}{3}}{\binom{100}{3}} = \frac{4}{161700}$.

$P(\text{all 3 identified as pure} | 0 \text{ impure}) = (0.99)^3 = 0.970299$.

$P(\text{all 3 identified as pure} | 1 \text{ impure}) = (0.99)^2 (0.05) = 0.049005$.

$P(\text{all 3 identified as pure} | 2 \text{ impure}) = (0.99) (0.05)^2 = 0.002475$.

$P(\text{all 3 identified as pure} | 3 \text{ impure}) = (0.05)^3 = 0.000125$.

$P(\text{accept}) = (\frac{142880}{161700})(0.970299) + (\frac{18240}{161700})(0.049005) + (\frac{576}{161700})(0.002475) + (\frac{4}{161700})(0.000125) = 0.857613 + 0.005524 + 0.0000088 + 0.0000000031 = 0.863146$.

3. Final Answer

The probability that this batch of instruments will be accepted is approximately 0.

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