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Probability question

2023 · 6 Apr · Shift 2 · Q34
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  5. /2023 · 6 Apr · Shift 2 · Q34

Probability question

2023 · 6 Apr · Shift 2 · Q34

JEE MainMathematicsProbabilityMCQ+4 / −1
Three dice are rolled. If the probability of getting different numbers on the three dice is pq\frac{p}{q}qp​, where ppp and qqq are co-prime, then q−pq-pq−p is equal to :
  1. A
    3
  2. B
    4
  3. C
    1
  4. D
    2
View written solutionFree

Correct answer: B

  1. Total number of outcomes

When three dice are rolled, each die has 666 possible outcomes. So total outcomes are:

63=2166^3 = 21663=216

  1. Favourable outcomes: all three dice show different numbers

We want all three numbers to be distinct.

  • First die: 666 choices
  • Second die: must be different from first  5$ choices
  • Third die: must be different from first two  4$ choices

So favourable outcomes are:

6×5×4=1206 \times 5 \times 4 = 1206×5×4=120

  1. Required probability

P(all different)=120216P(\text{all different}) = \frac{120}{216}P(all different)=216120​

Simplify:

120216=59\frac{120}{216} = \frac{5}{9}216120​=95​

Thus,

p=5,q=9p=5, \quad q=9p=5,q=9

  1. Find q−pq-pq−p

q−p=9−5=4q-p = 9-5 = 4q−p=9−5=4

  1. Option check
  • A: 333  Incorrect
  • B: 444  Correct
  • C: 111  Incorrect
  • D: 222  Incorrect

Therefore, the correct answer is B.

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