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Probability question

2020 · 3 Sep · Shift 1 · Q36
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Probability question

2020 · 3 Sep · Shift 1 · Q36

JEE MainMathematicsProbabilityMCQ+4 / −1
A dice is thrown two times and the sum of the scores appearing on the die is observed to be a multiple of 4. Then the conditional probability that the score 4 has appeared atleast once is :
  1. A
    18{1 \over 8}81​
  2. B
    19{1 \over 9}91​
  3. C
    14{1 \over 4}41​
  4. D
    13{1 \over 3}31​
View written solutionFree

Correct answer: B

  1. Define the events

Let:

  • AAA = “the sum of the two throws is a multiple of 444”
  • BBB = “the score 444 appears at least once”

We need to find: P(B∣A)=P(A∩B)P(A)P(B\mid A)=\frac{P(A\cap B)}{P(A)}P(B∣A)=P(A)P(A∩B)​

Since a die is thrown twice, total possible outcomes are: 6×6=366\times 6=366×6=36

  1. Find event AAA: sum is a multiple of 444

Possible sums that are multiples of 444 are: 4,8,124,8,124,8,12

Now list the ordered pairs:

  • Sum 444: (1,3),(2,2),(3,1)(1,3),(2,2),(3,1)(1,3),(2,2),(3,1)
    Total = 333

  • Sum 888: (2,6),(3,5),(4,4),(5,3),(6,2)(2,6),(3,5),(4,4),(5,3),(6,2)(2,6),(3,5),(4,4),(5,3),(6,2)
    Total = 555

  • Sum 121212: (6,6)(6,6)(6,6)
    Total = 111

Hence, n(A)=3+5+1=9n(A)=3+5+1=9n(A)=3+5+1=9

So, P(A)=936=14P(A)=\frac{9}{36}=\frac{1}{4}P(A)=369​=41​

  1. Find event A∩BA\cap BA∩B: sum is a multiple of 444 and at least one 444 appears

We check the outcomes in AAA that contain at least one 444.

From the list above:

  • For sum 444: no outcome contains 444
  • For sum 888: only (4,4)(4,4)(4,4) contains 444
  • For sum 121212: (6,6)(6,6)(6,6) does not contain 444

So the only favorable outcome is: (4,4)(4,4)(4,4)

Thus, n(A∩B)=1n(A\cap B)=1n(A∩B)=1

Therefore, P(A∩B)=136P(A\cap B)=\frac{1}{36}P(A∩B)=361​

  1. Compute the conditional probability
=\frac{\frac{1}{36}}{\frac{9}{36}} =\frac{1}{9}$$ 5. **Evaluate the options** - A: $\frac{1}{8}$ ❌ - B: $\frac{1}{9}$ ✅ - C: $\frac{1}{4}$ ❌ - D: $\frac{1}{3}$ ❌ Hence, the correct answer is: $$\boxed{\frac{1}{9}}$$
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