Example 29 - Chapter 13 Class 12 Probability

Transcript

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 – 4) 48 cards = 48C2 = 1128 P(X = 0) = 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑡𝑜 𝑔𝑒𝑡 0 𝑘𝑖𝑛𝑔𝑠﷮𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠﷯ = 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 × Number of ways to select 1 card from 48 non king cards = 4C1 × 48C1 = 4 × 48 = 192 P(X = 1) = 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑡𝑜 𝑔𝑒𝑡 1 𝑘𝑖𝑛𝑔﷮𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠﷯ = 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) = 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑡𝑜 𝑔𝑒𝑡 2 𝑘𝑖𝑛𝑔𝑠﷮𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠﷯ = 6﷮1326﷯ The probability distribution is The expectation value E(x) is given by 𝜇=E(X)= 𝑖 = 1﷮𝑛﷮𝑥𝑖𝑝𝑖﷯ = 0 × 1128﷮1326﷯ +1 × 192﷮1326﷯ + 2 × 6﷮1326﷯ = 0 + 192 + 12 ﷮1326﷯ = 204﷮1326﷯ = 34﷮221﷯ = 2﷮13﷯ The variance of x is given by : Var 𝑋﷯=𝐸 𝑋﷮2﷯﷯− 𝐸 𝑋﷯﷯﷮2﷯ So, finding 𝐸 𝑋﷮2﷯﷯ E 𝑋﷮2﷯﷯= 𝑖 = 1﷮𝑛﷮ 𝑥﷮𝑖﷯﷮2﷯𝑝𝑖﷯ = 02 × 1128﷮1326﷯+12 × 192﷮1326﷯+ 22 × 6﷮1326﷯ = 0+ 192 + 4 × 6﷮1326﷯ = 192 + 24﷮1326﷯ = 216﷮1326﷯ = 36﷮221﷯ Now, Var 𝑿﷯=𝑬 𝑿﷮𝟐﷯﷯− 𝑬 𝑿﷯﷯﷮𝟐﷯ = 36﷮221﷯− 34﷮221﷯﷯﷮2﷯ = 1﷮221﷯ 36− 34﷮2﷯﷮221﷯﷯ = 1﷮221﷯ 221 × 36 − 1156﷮221﷯﷯ = 6800﷮ 221﷯﷮2﷯﷯ ∴ Variance var 𝑋﷯ = 6800﷮ 221﷯﷮2﷯﷯ Standard deviation is given by 𝝈﷮𝒙﷯= ﷮𝑣𝑎𝑟 𝑋﷯﷯ = ﷮ 6800﷮ 221﷯﷮2﷯﷯﷯ = ﷮6800﷯﷮221﷯ = 0.37