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

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  1. Chapter 13 Class 12 Probability
  2. Serial order wise
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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. 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 โ€“ 4) 48 cards = 48C2 = 1128 P(X = 0) = ๐‘๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘ค๐‘Ž๐‘ฆ๐‘  ๐‘ก๐‘œ ๐‘”๐‘’๐‘ก 0 ๐‘˜๐‘–๐‘›๐‘”๐‘ ๏ทฎ๐‘‡๐‘œ๐‘ก๐‘Ž๐‘™ ๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘ค๐‘Ž๐‘ฆ๐‘ ๏ทฏ = 1128๏ทฎ1326๏ทฏ 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 ร— Number of ways to select 1 card from 48 non king cards = 4C1 ร— 48C1 = 4 ร— 48 = 192 P(X = 1) = ๐‘๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘ค๐‘Ž๐‘ฆ๐‘  ๐‘ก๐‘œ ๐‘”๐‘’๐‘ก 1 ๐‘˜๐‘–๐‘›๐‘”๏ทฎ๐‘‡๐‘œ๐‘ก๐‘Ž๐‘™ ๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘ค๐‘Ž๐‘ฆ๐‘ ๏ทฏ = 192๏ทฎ1326๏ทฏ 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) = ๐‘๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘ค๐‘Ž๐‘ฆ๐‘  ๐‘ก๐‘œ ๐‘”๐‘’๐‘ก 2 ๐‘˜๐‘–๐‘›๐‘”๐‘ ๏ทฎ๐‘‡๐‘œ๐‘ก๐‘Ž๐‘™ ๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘ค๐‘Ž๐‘ฆ๐‘ ๏ทฏ = 6๏ทฎ1326๏ทฏ The probability distribution is The expectation value E(x) is given by ๐œ‡=E(X)= ๐‘– = 1๏ทฎ๐‘›๏ทฎ๐‘ฅ๐‘–๐‘๐‘–๏ทฏ = 0 ร— 1128๏ทฎ1326๏ทฏ +1 ร— 192๏ทฎ1326๏ทฏ + 2 ร— 6๏ทฎ1326๏ทฏ = 0 + 192 + 12 ๏ทฎ1326๏ทฏ = 204๏ทฎ1326๏ทฏ = 34๏ทฎ221๏ทฏ = 2๏ทฎ13๏ทฏ The variance of x is given by : Var ๐‘‹๏ทฏ=๐ธ ๐‘‹๏ทฎ2๏ทฏ๏ทฏโˆ’ ๐ธ ๐‘‹๏ทฏ๏ทฏ๏ทฎ2๏ทฏ So, finding ๐ธ ๐‘‹๏ทฎ2๏ทฏ๏ทฏ E ๐‘‹๏ทฎ2๏ทฏ๏ทฏ= ๐‘– = 1๏ทฎ๐‘›๏ทฎ ๐‘ฅ๏ทฎ๐‘–๏ทฏ๏ทฎ2๏ทฏ๐‘๐‘–๏ทฏ = 02 ร— 1128๏ทฎ1326๏ทฏ+12 ร— 192๏ทฎ1326๏ทฏ+ 22 ร— 6๏ทฎ1326๏ทฏ = 0+ 192 + 4 ร— 6๏ทฎ1326๏ทฏ = 192 + 24๏ทฎ1326๏ทฏ = 216๏ทฎ1326๏ทฏ = 36๏ทฎ221๏ทฏ Now, Var ๐‘ฟ๏ทฏ=๐‘ฌ ๐‘ฟ๏ทฎ๐Ÿ๏ทฏ๏ทฏโˆ’ ๐‘ฌ ๐‘ฟ๏ทฏ๏ทฏ๏ทฎ๐Ÿ๏ทฏ = 36๏ทฎ221๏ทฏโˆ’ 34๏ทฎ221๏ทฏ๏ทฏ๏ทฎ2๏ทฏ = 1๏ทฎ221๏ทฏ 36โˆ’ 34๏ทฎ2๏ทฏ๏ทฎ221๏ทฏ๏ทฏ = 1๏ทฎ221๏ทฏ 221 ร— 36 โˆ’ 1156๏ทฎ221๏ทฏ๏ทฏ = 6800๏ทฎ 221๏ทฏ๏ทฎ2๏ทฏ๏ทฏ โˆด Variance var ๐‘‹๏ทฏ = 6800๏ทฎ 221๏ทฏ๏ทฎ2๏ทฏ๏ทฏ Standard deviation is given by ๐ˆ๏ทฎ๐’™๏ทฏ= ๏ทฎ๐‘ฃ๐‘Ž๐‘Ÿ ๐‘‹๏ทฏ๏ทฏ = ๏ทฎ 6800๏ทฎ 221๏ทฏ๏ทฎ2๏ทฏ๏ทฏ๏ทฏ = ๏ทฎ6800๏ทฏ๏ทฎ221๏ทฏ = 0.37

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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