Question 4 - Case Based Questions (MCQ) - Chapter 13 Class 12 Probability

Last updated at May 29, 2023 by

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:

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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 βˆ’ 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 – 4 = ) 48 cards = 48C2 = 48!/(2! (48 βˆ’ 2)!) = 48!/(2! (46)!) = 1128 P(X = 0) = (π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘€π‘Žπ‘¦π‘  π‘‘π‘œ 𝑔𝑒𝑑 0 π‘˜π‘–π‘›π‘”π‘ )/(π‘‡π‘œπ‘‘π‘Žπ‘™ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘€π‘Žπ‘¦π‘ ) = 1128/1326 = πŸπŸ–πŸ–/𝟐𝟐𝟏 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‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘€π‘Žπ‘¦π‘  π‘‘π‘œ 𝑔𝑒𝑑 1 π‘˜π‘–π‘›π‘”)/(π‘‡π‘œπ‘‘π‘Žπ‘™ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘€π‘Žπ‘¦π‘ ) = 192/1326 = πŸ‘πŸ/𝟐𝟐𝟏 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‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘€π‘Žπ‘¦π‘  π‘‘π‘œ 𝑔𝑒𝑑 2 π‘˜π‘–π‘›π‘”π‘ )/(π‘‡π‘œπ‘‘π‘Žπ‘™ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘€π‘Žπ‘¦π‘ ) = 6/1326 = 𝟏/𝟐𝟐𝟏 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œ‡="E(X)"=βˆ‘2_(𝑖 = 1)^𝑛▒𝑋𝑖𝑃𝑖 = 0 Γ— 188/221 +"1 Γ—" 32/221 + 2 Γ— 1/221 = 0 + (32+2 )/221 = 34/221 = 𝟐/πŸπŸ‘ 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 (𝑋)=𝐸(𝑋^2 )βˆ’[𝐸(𝑋)]^2 Finding 𝑬(𝑿^𝟐 ) E(𝑿^𝟐 )=βˆ‘2_(𝑖 = 1)^𝑛▒〖〖π‘₯_𝑖〗^2 𝑝𝑖〗 = 02 Γ— 188/221+"12 Γ— " 32/221+ 22 Γ— 1/221= 0+(32 + 4 )/221 = πŸ‘πŸ”/𝟐𝟐𝟏 Now, Var (𝑿)=𝑬(𝑿^𝟐 )βˆ’[𝑬(𝑿)]^𝟐 = 36/221βˆ’(34/221)^2 = 1/221 [36βˆ’γ€–34γ€—^2/221] = 1/221 [(221 Γ— 36 βˆ’ 1156)/221] = 6800/(221)^2 = πŸ’πŸŽπŸŽ/πŸπŸ–πŸ•πŸ‘ So, the correct answer is (A)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.