# Ex 13.4, 17 - Chapter 13 Class 12 Probability

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 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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