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Chapter 13 Class 12 Probability

Serial order wise

Last updated at Feb. 15, 2020 by Teachoo

Example 29 Two cards are drawn simultaneously (or successively without replacement) from a well shuffled pack of 52 cards. Find the mean, variance and standard deviation of the number of kings.Since we are drawing cards without replacement, it is NOT a Bernoulli trial Let X be the number of kings obtained We can get 0, 1, or 2 kings 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 kings Number of ways to get 0 kings = Number of ways to select 2 cards out of non king 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 πππππ )/(πππ‘ππ ππ’ππππ ππ π€ππ¦π ) = 1128/1326 P(X = 1) i.e. Probability of getting 1 kings Number of ways to get 1 kings = Number of ways to select 1 king out of 4 king cards Γ Number of ways to select 1 card from 48 non king 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 kings Number of ways to get 1 kings = Number of ways of selecting 2 kings out of 4 king cards = 4C2 = 6 P(X = 2) = (ππ’ππππ ππ π€ππ¦π π‘π πππ‘ 2 πππππ )/(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 = ππ/πππ The variance of x is given by : Var (π)=πΈ(π^2 )β[πΈ(π)]^2 So, finding π¬(πΏ^π ) E(π^2 )=β2_(π = 1)^πβγγπ₯_πγ^2 ππγ = 02 Γ 1128/1326+"12 Γ " 192/1326+ 22 Γ 6/1326 = 0+(192 + 4 Γ 6)/1326 = (192 + 24)/1326 = 216/1326 = 36/221 Now, Var (πΏ)=π¬(πΏ^π )β[π¬(πΏ)]^π = 36/221β(34/221)^2 = 1/221 [36βγ34γ^2/221] = 1/221 [(221 Γ 36 β 1156)/221] = 6800/(221)^2 β΄ Variance var (π) = 6800/(221)^2 Standard deviation is given by π_π=β(π£ππ(π) ) =β(6800/(221)^2 ) = β6800/221 = 82.46/221 = 8246/22100 = 0.37