Let T = truth, F = false. P(T)=4/5, P(F)=1/5. P(actual 5)=1/6, P(not 5)=5/6. By Bayes' theorem: P(actual 5∣reports 5)=(4/5)(1/6)+(1/5)(5/6)(4/5)(1/6)=4/30+5/304/30=94.
A man is known to speak truth 4 out of 5 times. He throws a die and reports that five appears. Then the probability that actual five appears on the…
CUET UG 2022 — Mathematics Algebra
2022mcqmedium
A man is known to speak truth 4 out of 5 times. He throws a die and reports that five appears. Then the probability that actual five appears on the dice is
Official previous-year question
Held on 10 Aug 2022 · Verified 13 Jul 2026.
Options
- A
83
- B
94
- C
95
- D
85
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