Check sibling questions

Ex 13.4, 17 - Let X be number of aces. Then value of E(X) is

Ex 13.4, 17 - Chapter 13 Class 12 Probability - Part 2
Ex 13.4, 17 - Chapter 13 Class 12 Probability - Part 3
Ex 13.4, 17 - Chapter 13 Class 12 Probability - Part 4
Ex 13.4, 17 - Chapter 13 Class 12 Probability - Part 5


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) = (π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘€π‘Žπ‘¦π‘  π‘‘π‘œ 𝑔𝑒𝑑 0 π‘Žπ‘π‘’π‘ )/(π‘‡π‘œπ‘‘π‘Žπ‘™ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘€π‘Žπ‘¦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) = (π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘€π‘Žπ‘¦π‘  π‘‘π‘œ 𝑔𝑒𝑑 1 π‘Žπ‘π‘’π‘ )/(π‘‡π‘œπ‘‘π‘Žπ‘™ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘€π‘Žπ‘¦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) = (π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘€π‘Žπ‘¦π‘  π‘‘π‘œ 𝑔𝑒𝑑 2 π‘Žπ‘π‘’π‘ )/(π‘‡π‘œπ‘‘π‘Žπ‘™ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘€π‘Žπ‘¦π‘ ) = 6/1326 The probability distribution is The expectation value E(x) is given by 𝝁="E(X)"=βˆ‘2_(𝑖 = 1)^𝑛▒π‘₯𝑖𝑝𝑖 = 0 Γ— 1128/1326 +"1 Γ—" 192/1326 + 2 Γ— 6/1326 = 0 + (192 + 12 )/1326 = 204/1326 = 𝟐/πŸπŸ‘ Hence option D is correct

Davneet Singh's photo - Teacher, Engineer, Marketer

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.