Question:
A six faced biased dice is so loaded that in any of its toss getting any of even scores is equally likely and the probability of getting an even score is three times the probability of getting an odd score. This dice is tossed repeatedly until a six is obtained. If X denotes the number of tosses required, then \(\mathrm{P}(\mathrm{X}<4)\) is :
A six faced biased dice is so loaded that in any of its toss getting any of even scores is equally likely and the probability of getting an even score is three times the probability of getting an odd score. This dice is tossed repeatedly until a six is obtained. If X denotes the number of tosses required, then \(\mathrm{P}(\mathrm{X}<4)\) is :
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Find P(six) from the condition that P(even) = 3 P(odd), then use P(X < 4) = 1 - P(no six in first 3 tosses).
Updated On: Nov 12, 2026
- \(\frac{19}{3^{3}}\)
- \(\frac{37}{4^{3}}\)
- \(\frac{61}{4^{3}}\)
- \(\frac{397}{(12)^{3}}\)
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The Correct Option is B
Solution and Explanation
Let P(even) = 3 P(odd). Since P(even) + P(odd) = 1, P(even) = 3/4 and P(odd) = 1/4.
The three even scores 2, 4, 6 are equally likely, so P(six) = (3/4)/3 = 1/4 and P(not six) = 3/4.
X < 4 means a six appears within the first 3 tosses, which is the complement of no six in 3 tosses.
P(X < 4) = 1 - (3/4)3 = 1 - 27/64 = 37/64 = 37/43.
Answer: (2) \(\frac{37}{4^{3}}\)
The three even scores 2, 4, 6 are equally likely, so P(six) = (3/4)/3 = 1/4 and P(not six) = 3/4.
X < 4 means a six appears within the first 3 tosses, which is the complement of no six in 3 tosses.
P(X < 4) = 1 - (3/4)3 = 1 - 27/64 = 37/64 = 37/43.
Answer: (2) \(\frac{37}{4^{3}}\)
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