# Ex 13.4, 17

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 13.4, 17(Method 1) 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) 37221 (B) 513 (C) 113 (D) 213 Picking a card is a Bernoulli trial So, X has a binomial distribution P(X = x) = nCx 𝒒𝒏−𝒙 𝒑𝒙 Here, n = number of cards drawn = 2 p = Probability of getting ace = 452 = 113 q = 1 – p = 1 – 113 = 1213 Hence, ⇒ P(X = x) = 2Cx 𝟏𝟏𝟑𝒙 𝟏𝟐𝟏𝟑𝟐 − 𝒙 Since 2 cards are drawn, We can get 0 ace, 1 ace or 2 ace So, X can be 0, 1 , 2 Putting values in (1) P(X = 0) = 2C0 1130 12132 −0 = 1 × 1 × 12132 = 144169 P(X = 1) = 2C1 1131 12132 − 1 = 2 × 113 × 1213 = 24169 P(X = 2) = 2C2 1132 12132 − 2 = 1 × 1132 = 1169 The probability distribution is The expectation value E(x) is given by 𝝁=E(X)= 𝑖 = 1𝑛𝑥𝑖𝑝𝑖 = 0 × 144169 +1 × 24169 + 2 × 1169 = 0 + 24 + 2 169 = 26169 = 𝟐𝟏𝟑 Hence option D is correct Ex 13.4, 17(Method 2) 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) 37221 (B) 513 (C) 113 (D) 213 Let 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 𝑎𝑐𝑒𝑠𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠 = 11281326 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 𝑎𝑐𝑒𝑠𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠 = 1921326 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 𝑎𝑐𝑒𝑠𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠 = 61326 The probability distribution is The expectation value E(x) is given by 𝝁=E(X)= 𝑖 = 1𝑛𝑥𝑖𝑝𝑖 = 0 × 11281326 +1 × 1921326 + 2 × 61326 = 0 + 192 + 12 1326 = 2041326 = 𝟐𝟏𝟑 Hence option D is correct

Chapter 13 Class 12 Probability

Serial order wise

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CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .