Test Index

Probability Part 3

Section: Mathematics

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Question : 23 of 35

Marks: +1, -0

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

[4 Apr 2025 Shift 1]

$\;\frac{103}{182}$
$\;\frac{19}{26}$
$\;\frac{17}{26}$
$\;\frac{129}{182}$

Solution:

Total number of ways to form a

12

-person committee from

16

people is

\binom{16}{12}=1820

.

Let

k

be the number of engineers and

j

be the number of doctors selected.

The conditions are

k\ge 3

and

j\ge 1

.

Case

k=3

,

j=1

:

\binom{4}{3}\binom{2}{1}\binom{10}{8}=360

.

Case

k=3

,

j=2

:

\binom{4}{3}\binom{2}{2}\binom{10}{7}=480

.

Case

k=4

,

j=1

:

\binom{4}{4}\binom{2}{1}\binom{10}{7}=240

.

Case

k=4

,

j=2

:

\binom{4}{4}\binom{2}{2}\binom{10}{6}=210

.

Total favorable committees

=360+480+240+210=1290

.

Required probability

=\frac{1290}{1820}=\frac{129}{182}

.

Thus the correct option is

\frac{129}{182}

, i.e. option

3

.

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