Case Based Questions (MCQ)

Chapter 13 Class 12 Probability
Serial order wise

## Based on the above information answer theΒ  following:

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### Transcript

Question A group of people start playing cards. And as we know a well shuffled pack of cards contains a total of 52 cards. Then 2 cards are drawn simultaneously (or successively without replacement). Based on the above information answer the following: Total number of ways to draw 2 cards out of 52 is Total ways = 52C2 = 52!/(2! (52 β 2)!) = 52!/(2! (50)!) = 1326 Question 1 If X = no. of kings = 0, 1, 2. Then P(X = 0) =? (A) 188/221 (B) 198/223 (C) 197/290 (D) 187/221 P(X = 0) i.e. Probability of getting 0 kings Number of ways to get 0 kings = Number of ways to select 2 cards out of non king cards = Number of ways to select 2 cards out of (52 β 4 = ) 48 cards = 48C2 = 48!/(2! (48 β 2)!) = 48!/(2! (46)!) = 1128 P(X = 0) = (ππ’ππππ ππ π€ππ¦π  π‘π πππ‘ 0 πππππ )/(πππ‘ππ ππ’ππππ ππ π€ππ¦π ) = 1128/1326 = πππ/πππ So, the correct answer is (a) Question 2 If X = no. of kings = 0, 1, 2. Then P(X = 1) =? (A) 32/229 (B) 32/227 (C) 32/221 (D) 32/219 P(X = 1) i.e. Probability of getting 1 king Number of ways to get 1 king = Number of ways to select 1 king out of 4 king cards Γ Number of ways to select 1 card from 48 non king cards = 4C1 Γ 48C1 = 4 Γ 48 = 192 P(X = 1) = (ππ’ππππ ππ π€ππ¦π  π‘π πππ‘ 1 ππππ)/(πππ‘ππ ππ’ππππ ππ π€ππ¦π ) = 192/1326 = ππ/πππ So, the correct answer is (c) Question 3 If X = no. of kings = 0, 1, 2. Then P(X = 2) =? (A) 2/219 (B) 1/221 (C) 3/209 (D) 1/209 P(X = 2) i.e. Probability of getting 2 kings Number of ways to get 2 kings = Number of ways of selecting 2 kings out of 4 king cards = 4C2 = 4!/(2! 2!) = 6 P(X = 2) = (ππ’ππππ ππ π€ππ¦π  π‘π πππ‘ 2 πππππ )/(πππ‘ππ ππ’ππππ ππ π€ππ¦π ) = 6/1326 = π/πππ So, the correct answer is (b) Question 4 Find the mean of the number of Kings? (A) 2/13 (B) 1/13 (C) 1/17 (D) 2/17 The probability distribution is The expectation value E(X) is given by π="E(X)"=β2_(π = 1)^πβππππ = 0 Γ 188/221 +"1 Γ" 32/221 + 2 Γ 1/221 = 0 + (32+2 )/221 = 34/221 = π/ππ So, the correct answer is (a) Question 5 Find the variance of the number of Kings? (A) 400/2873 (B) 400/2877 (C) 400/2879 (D) 400/2871 The variance of x is given by : Var (π)=πΈ(π^2 )β[πΈ(π)]^2 Finding π¬(πΏ^π ) E(πΏ^π )=β2_(π = 1)^πβγγπ₯_πγ^2 ππγ = 02 Γ 188/221+"12 Γ " 32/221+ 22 Γ 1/221= 0+(32 + 4 )/221 = ππ/πππ Now, Var (πΏ)=π¬(πΏ^π )β[π¬(πΏ)]^π = 36/221β(34/221)^2 = 1/221 [36βγ34γ^2/221] = 1/221 [(221 Γ 36 β 1156)/221] = 6800/(221)^2 = πππ/ππππ So, the correct answer is (A)