Check sibling questions

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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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.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) = (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)"=∑2_(𝑖 = 1)^𝑛▒𝑥𝑖𝑝𝑖 = 0 × 1128/1326 +"1 ×" 192/1326 + 2 × 6/1326 = 0 + (192 + 12 )/1326 = 204/1326 = 𝟑𝟒/𝟐𝟐𝟏 The variance of x is given by : Var (𝑋)=𝐸(𝑋^2 )−[𝐸(𝑋)]^2 So, finding 𝑬(𝑿^𝟐 ) E(𝑋^2 )=∑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 = 82.46/221 = 8246/22100 = 0.37

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.