# Ex 13.4, 17 - Chapter 13 Class 12 Probability

Last updated at Feb. 15, 2020 by Teachoo

Last updated at Feb. 15, 2020 by Teachoo

Transcript

Ex 13.4, 17 Suppose that two cards are drawn at random from a deck of cards. Let X be the number of aces obtained. Then the value of E(X) is (A) 37/221 (B) 5/13 (C) 1/13 (D) 2/13Let X be the number of aces obtained We can get 0, 1, or 2 aces So, value of X is 0, 1 or 2 Total number of ways to draw 2 cards out of 52 is Total ways = 52C2 = 1326 P(X = 0) i.e. probability of getting 0 aces Number of ways to get 0 aces = Number of ways to select 2 cards out of non ace cards = Number of ways to select 2 cards out of (52 – 4) 48 cards = 48C2 = 1128 P(X = 0) = (𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑡𝑜 𝑔𝑒𝑡 0 𝑎𝑐𝑒𝑠)/(𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠) = 1128/1326 P(X = 1) i.e. probability of getting 1 aces Number of ways to get 1 aces = Number of ways to select 1 ace out of 4 ace cards × Number of ways to select 1 card from 48 non ace cards = 4C1 × 48C1 = 4 × 48 = 192 P(X = 1) = (𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑡𝑜 𝑔𝑒𝑡 1 𝑎𝑐𝑒𝑠)/(𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠) = 192/1326 P(X = 2) i.e. probability of getting 2 aces Number of ways to get 1 aces = Number of ways of selecting 2 aces out of 4 ace cards = 4C2 = 6 P(X = 2) = (𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑡𝑜 𝑔𝑒𝑡 2 𝑎𝑐𝑒𝑠)/(𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠) = 6/1326 The probability distribution is The expectation value E(x) is given by 𝝁="E(X)"=∑2_(𝑖 = 1)^𝑛▒𝑥𝑖𝑝𝑖 = 0 × 1128/1326 +"1 ×" 192/1326 + 2 × 6/1326 = 0 + (192 + 12 )/1326 = 204/1326 = 𝟐/𝟏𝟑 Hence option D is correct

Ex 13.4

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Ex 13.4, 3 Important

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Ex 13.4, 6 Important

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Ex 13.4, 10 Deleted for CBSE Board 2021 Exams only

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Ex 13.4, 12 Important Deleted for CBSE Board 2021 Exams only

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Ex 13.4, 15 Deleted for CBSE Board 2021 Exams only

Ex 13.4, 16 Deleted for CBSE Board 2021 Exams only

Ex 13.4, 17 Important Deleted for CBSE Board 2021 Exams only You are here

Chapter 13 Class 12 Probability

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.