



Mean or Expectation of random variable
Ex 13.4, 16 (MCQ) Deleted for CBSE Board 2022 Exams
Ex 13.4, 11 Important Deleted for CBSE Board 2022 Exams
Example 27 Deleted for CBSE Board 2022 Exams
Ex 13.4, 12 Important Deleted for CBSE Board 2022 Exams
Ex 13.4, 17 (MCQ) Important Deleted for CBSE Board 2022 Exams You are here
Mean or Expectation of random variable
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 β 4) 48 cards = 48C2 = 1128 P(X = 0) = (ππ’ππππ ππ π€ππ¦π π‘π πππ‘ 0 ππππ )/(πππ‘ππ ππ’ππππ ππ π€ππ¦π ) = 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 Γ Number of ways to select 1 card from 48 non ace cards = 4C1 Γ 48C1 = 4 Γ 48 = 192 P(X = 1) = (ππ’ππππ ππ π€ππ¦π π‘π πππ‘ 1 ππππ )/(πππ‘ππ ππ’ππππ ππ π€ππ¦π ) = 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) = (ππ’ππππ ππ π€ππ¦π π‘π πππ‘ 2 ππππ )/(πππ‘ππ ππ’ππππ ππ π€ππ¦π ) = 6/1326 The probability distribution is The expectation value E(x) is given by π="E(X)"=β2_(π = 1)^πβπ₯πππ = 0 Γ 1128/1326 +"1 Γ" 192/1326 + 2 Γ 6/1326 = 0 + (192 + 12 )/1326 = 204/1326 = π/ππ Hence option D is correct