The probability, of forming a 12 persons committee from 4 engineers, 2 doctors and 10 professors containing at least 3 engineers and at least 1 doctor, is [2025]

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Solution

Q.
The probability, of forming a 12 persons committee from 4 engineers, 2 doctors and 10 professors containing at least 3 engineers and at least 1 doctor, is [2025]

1 $\frac{129}{182}$

2 $\frac{103}{182}$

3 $\frac{17}{26}$

4 $\frac{19}{26}$

Ans.
(1)

Total number of ways of selecting 12 people from 4 engineers, 2 doctors and 10 professors = ${}^{16}C_{12}$ = 1820.

To determine favourite outcomes:-

To select at least 3 engineers 1 doctor is, we can choose

4 Engineers 2 Doctors 10 Professors Number of ways

3 1 8 ${}^{4}C_{3} \times {}^{2}C_{1} \times {}^{10}C_{8} = 360$

4 2 6 ${}^{4}C_{4} \times {}^{2}C_{2} \times {}^{10}C_{6} = 210$

3 2 7 ${}^{4}C_{3} \times {}^{2}C_{2} \times {}^{10}C_{7} = 480$

4 1 7 ${}^{4}C_{4} \times {}^{2}C_{1} \times {}^{10}C_{7} = 240$

Total favourable outcomes = 1290

So, required probability $= \frac{1290}{1820} = \frac{129}{182}$ .

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