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Question 30 of 48

Q.(a) A cricket team of 11 players is to be formed from 16 players including 4 bowlers and 2 wicket-keepers. In how many different ways can a team be formed so that the team contains atleast 3 bowlers and atleast one wicket-keeper ?

(OR)
(b) X speaks truth 4 out of 5 times. A die is thrown. He reports that there is a six. What is the chance that actually there was a six ?
Tamil Nadu DgeTamil Nadu HSC First Year (DGE) Commerce Board 2022Subjective· 5mImportance★★★★★
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(a) Sum the case counts for ≥3\ge3 bowlers and ≥1\ge1 wicket-keeper: 24722472 ways. (b) Bayes' theorem gives 49\dfrac{4}{9}.

Part (a) — Cricket team selection.

The 16 players are 4 bowlers (BB), 2 wicket-keepers (WW) and 16−4−2=1016-4-2=10 others (OO). We need 11 players with at least 3 bowlers and at least 1 wicket-keeper. Enumerate the allowed combinations:

BowlersW-keepersOthersCount
3 ((43)=4\binom{4}{3}=4)1 ((21)=2\binom{2}{1}=2)7 ((107)=120\binom{10}{7}=120)4×2×120=9604\times2\times120=960
3 (44)2 ((22)=1\binom{2}{2}=1)6 ((106)=210\binom{10}{6}=210)4×1×210=8404\times1\times210=840
4 ((44)=1\binom{4}{4}=1)1 (22)6 (210210)1×2×210=4201\times2\times210=420
4 (11)2 (11)5 ((105)=252\binom{10}{5}=252)1×1×252=2521\times1\times252=252

Total=960+840+420+252=2472.\text{Total}=960+840+420+252=2472.

Part (b) — Truth-telling and Bayes' theorem.

Let S=S= "a six actually turns up", T=T= "X reports a six".

P(S)=16, P(S′)=56,P(X speaks truth)=45, P(X lies)=15.P(S)=\dfrac16,\ P(S')=\dfrac56,\qquad P(\text{X speaks truth})=\dfrac45,\ P(\text{X lies})=\dfrac15. …

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