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Chapter 13 Class 12 Probability

Example 29 - Find mean, variance, standard deviation of kings

Example 29 - Chapter 13 Class 12 Probability - Part 2
Example 29 - Chapter 13 Class 12 Probability - Part 3 Example 29 - Chapter 13 Class 12 Probability - Part 4 Example 29 - Chapter 13 Class 12 Probability - Part 5 Example 29 - Chapter 13 Class 12 Probability - Part 6 Example 29 - Chapter 13 Class 12 Probability - Part 7 Example 29 - Chapter 13 Class 12 Probability - Part 8

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Question 8 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 – 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‘˜π‘–π‘›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 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) = (π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘€π‘Žπ‘¦π‘  π‘‘π‘œ 𝑔𝑒𝑑 1 π‘˜π‘–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 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) = (π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘€π‘Žπ‘¦π‘  π‘‘π‘œ 𝑔𝑒𝑑 2 π‘˜π‘–π‘›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 πœ‡="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 = πŸ‘πŸ’/𝟐𝟐𝟏 The variance of x is given by : Var (𝑋)=𝐸(𝑋^2 )βˆ’[𝐸(𝑋)]^2 So, finding 𝑬(𝑿^𝟐 ) E(𝑋^2 )=βˆ‘2_(𝑖 = 1)^𝑛▒〖〖π‘₯_𝑖〗^2 𝑝𝑖〗 = 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 (𝑿)=𝑬(𝑿^𝟐 )βˆ’[𝑬(𝑿)]^𝟐 = 36/221βˆ’(34/221)^2 = 1/221 [36βˆ’γ€–34γ€—^2/221] = 1/221 [(221 Γ— 36 βˆ’ 1156)/221] = 6800/(221)^2 ∴ Variance var (𝑋) = 6800/(221)^2 Standard deviation is given by 𝝈_𝒙=√(π‘£π‘Žπ‘Ÿ(𝑋) ) =√(6800/(221)^2 ) = √6800/221 = 82.46/221 = 8246/22100 = 0.37

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.