Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Probability question

2025 · 4 Apr · Shift 1 · Q36
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Probability
  5. /2025 · 4 Apr · Shift 1 · Q36

Probability question

2025 · 4 Apr · Shift 1 · Q36

JEE MainMathematicsProbabilityMCQ+4 / −1
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
  1. A
    129182\frac{129}{182}182129​
  2. B
    1726\frac{17}{26}2617​
  3. C
    1926\frac{19}{26}2619​
  4. D
    103182\frac{103}{182}182103​
View written solutionFree

Correct answer: A

  1. Interpret the question

We have a total of:

  • 444 engineers
  • 222 doctors
  • 101010 professors

So total people =4+2+10=16= 4+2+10 = 16=4+2+10=16.

A committee of 121212 persons is to be formed.

We need the probability that the committee contains:

  • at least 333 engineers
  • at least 111 doctor
  1. Total number of ways to form a 12-person committee
Total ways=(1612)\text{Total ways} = {16 \choose 12}Total ways=(1216​)

Using symmetry,

(1612)=(164)=1820{16 \choose 12} = {16 \choose 4} = 1820(1216​)=(416​)=1820
  1. Count favorable committees

Let the committee contain:

  • eee engineers,
  • ddd doctors,
  • ppp professors.

Then

e+d+p=12e+d+p=12e+d+p=12

Subject to:

  • e≥3e \ge 3e≥3, with at most 444 engineers available, so e=3e=3e=3 or 444
  • d≥1d \ge 1d≥1, with at most 222 doctors available, so d=1d=1d=1 or 222

Now check possible cases.

Case 1: e=3e=3e=3

Then

d+p=9d+p=9d+p=9

Since d=1d=1d=1 or 222:

Subcase 1.1: e=3,d=1e=3, d=1e=3,d=1

Then

p=8p=8p=8

Number of committees:

(43)(21)(108)=4⋅2⋅45=360{4 \choose 3}{2 \choose 1}{10 \choose 8} = 4 \cdot 2 \cdot 45 = 360(34​)(12​)(810​)=4⋅2⋅45=360

Subcase 1.2: e=3,d=2e=3, d=2e=3,d=2

Then

p=7p=7p=7

Number of committees:

(43)(22)(107)=4⋅1⋅120=480{4 \choose 3}{2 \choose 2}{10 \choose 7} = 4 \cdot 1 \cdot 120 = 480(34​)(22​)(710​)=4⋅1⋅120=480

So total for Case 1:

360+480=840360+480=840360+480=840

Case 2: e=4e=4e=4

Then

d+p=8d+p=8d+p=8

Again d=1d=1d=1 or 222.

Subcase 2.1: e=4,d=1e=4, d=1e=4,d=1

Then

p=7p=7p=7

Number of committees:

(44)(21)(107)=1⋅2⋅120=240{4 \choose 4}{2 \choose 1}{10 \choose 7} = 1 \cdot 2 \cdot 120 = 240(44​)(12​)(710​)=1⋅2⋅120=240

Subcase 2.2: e=4,d=2e=4, d=2e=4,d=2

Then

p=6p=6p=6

Number of committees:

(44)(22)(106)=1⋅1⋅210=210{4 \choose 4}{2 \choose 2}{10 \choose 6} = 1 \cdot 1 \cdot 210 = 210(44​)(22​)(610​)=1⋅1⋅210=210

So total for Case 2:

240+210=450240+210=450240+210=450
  1. Total favorable committees
840+450=1290840+450=1290840+450=1290
  1. Required probability
P=12901820P = \frac{1290}{1820}P=18201290​

Simplify:

12901820=129182\frac{1290}{1820} = \frac{129}{182}18201290​=182129​
  1. Compare with options

The probability is

129182\boxed{\frac{129}{182}}182129​​

So the correct option is A.

  1. Comparison with stored correct answer

Stored correct answer: A

Our derived answer is also A, so they agree.

PreviousNext

More from Probability

  • A box contains 10 pens of which 3 are defective. A sample of 2 pens is drawn at random and let X denote the number of defective pens. Then the variance of X is2025 · MCQ
  • A card from a pack of 52 cards is lost. From the remaining 51 cards, n cards are drawn and are found to be spades. If the probability of the lost card to be a spade is 5011​, then n is equal to ​ .2025 · Numerical
  • A bag contains 19 unbiased coins and one coin with head on both sides. One coin drawn at random is tossed and head turns up. If the probability that the drawn coin was unbiased, is nm​, gcd(m,n)=1, then n2−m2 is equal…2025 · MCQ
  • Let a random variable X take values 0, 1, 2, 3 with P(X=0)=P(X=1)=p, P(X=2)=P(X=3) and E(X2)=2E(X). Then the value of 8p−1 is :2025 · MCQ
  • If A and B are two events such that P(A)=0.7, P(B)=0.4 and P(A∩B)=0.5, where B denotes the complement of B, then P(B∣(A∪B)) is equal to2025 · MCQ
  • A coin is tossed three times. Let X denote the number of times a tail follows a head. If μ and σ2 denote the mean and variance of X, then the value of 64(μ+σ2) is:2025 · MCQ
  • Two balls are selected at random one by one without replacement from a bag containing 4 white and 6 black balls. If the probability that the first selected ball is black, given that the second selected ball is also black, is nm​,…2025 · MCQ
  • If A and B are two events such that P(A∩B)=0.1, and P(A∣B) and P(B∣A) are the roots of the equation 12x2−7x+1=0, then the value of P(Aˉ∩Bˉ)P(Aˉ∪Bˉ)​ is :2025 · MCQ