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Probability question

2009 · Shift 1 · Q27
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  5. /2009 · Shift 1 · Q27

Probability question

2009 · Shift 1 · Q27

JEE AdvancedMathematicsProbabilityMCQ+4 / −1
A fair die is tossed repeatedly until a six is obtained. Let X denote the number of tosses required.The probability that X = 3 equals
  1. A
    25216{{25} \over {216}}21625​
  2. B
    2536{{25} \over {36}}3625​
  3. C
    536{{5} \over {36}}365​
  4. D
    125216{{125} \over {216}}216125​
View written solutionFree

Correct answer: A

  1. Let XXX be the number of tosses required to get the first six.

  2. For X=3X=3X=3, the following must happen:

    • First toss is not a six
    • Second toss is not a six
    • Third toss is a six

  3. Since the die is fair, P(not six)=56,P(six)=16P(\text{not six})=\frac{5}{6}, \qquad P(\text{six})=\frac{1}{6}P(not six)=65​,P(six)=61​

  4. These tosses are independent, so P(X=3)=56⋅56⋅16P(X=3)=\frac{5}{6}\cdot \frac{5}{6}\cdot \frac{1}{6}P(X=3)=65​⋅65​⋅61​

  5. Multiply: P(X=3)=2536⋅16=25216P(X=3)=\frac{25}{36}\cdot \frac{1}{6}=\frac{25}{216}P(X=3)=3625​⋅61​=21625​

  6. Comparing with the options:

    • A: 25216\frac{25}{216}21625​ ✅
    • B: 2536\frac{25}{36}3625​
    • C: 536\frac{5}{36}365​
    • D: 125216\frac{125}{216}216125​

Hence, the correct answer is A.

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