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

Last updated at April 16, 2024 by Teachoo

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:

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 − 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 − 2)!)
= 48!/(2! (46)!)
= 1128
P(X = 0) = (𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑡𝑜 𝑔𝑒𝑡 0 𝑘𝑖𝑛𝑔𝑠)/(𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠)
= 1128/1326
= 𝟏𝟖𝟖/𝟐𝟐𝟏
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 × Number of ways to select 1 card from 48 non king cards
= 4C1 × 48C1
= 4 × 48
= 192
P(X = 1) = (𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑡𝑜 𝑔𝑒𝑡 1 𝑘𝑖𝑛𝑔)/(𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠)
= 192/1326
= 𝟑𝟐/𝟐𝟐𝟏
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) = (𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑡𝑜 𝑔𝑒𝑡 2 𝑘𝑖𝑛𝑔𝑠)/(𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠)
= 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
𝜇="E(X)"=∑2_(𝑖 = 1)^𝑛▒𝑋𝑖𝑃𝑖
= 0 × 188/221 +"1 ×" 32/221 + 2 × 1/221
= 0 + (32+2 )/221
= 34/221
= 𝟐/𝟏𝟑
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 )−[𝐸(𝑋)]^2
Finding 𝑬(𝑿^𝟐 )
E(𝑿^𝟐 )=∑2_(𝑖 = 1)^𝑛▒〖〖𝑥_𝑖〗^2 𝑝𝑖〗
= 02 × 188/221+"12 × " 32/221+ 22 × 1/221= 0+(32 + 4 )/221
= 𝟑𝟔/𝟐𝟐𝟏
Now,
Var (𝑿)=𝑬(𝑿^𝟐 )−[𝑬(𝑿)]^𝟐
= 36/221−(34/221)^2
= 1/221 [36−〖34〗^2/221]
= 1/221 [(221 × 36 − 1156)/221]
= 6800/(221)^2
= 𝟒𝟎𝟎/𝟐𝟖𝟕𝟑
So, the correct answer is (A)

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo

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