# Question 8 - Examples - Chapter 13 Class 12 Probability

Last updated at April 16, 2024 by Teachoo

Examples

Example 1

Example 2

Example 3

Example 4

Example 5 Important

Example 6

Example 7 Important

Example 8

Example 9 Important

Example 10

Example 11 Important

Example 12 Important

Example 13 Important

Example 14 Important

Example 15 Important

Example 16

Example 17 Important

Example 18 Important

Example 19 Important

Example 20 Important

Example 21 Important

Example 22 Important

Example 23 Important

Example 24 Important

Question 1 Deleted for CBSE Board 2024 Exams

Question 2 Deleted for CBSE Board 2024 Exams

Question 3 Important Deleted for CBSE Board 2024 Exams

Question 4 Important Deleted for CBSE Board 2024 Exams

Question 5 Important Deleted for CBSE Board 2024 Exams

Question 6 Deleted for CBSE Board 2024 Exams

Question 7 Important Deleted for CBSE Board 2024 Exams

Question 8 Important Deleted for CBSE Board 2024 Exams You are here

Question 9 Important Deleted for CBSE Board 2024 Exams

Question 10 Important Deleted for CBSE Board 2024 Exams

Question 11 Important Deleted for CBSE Board 2024 Exams

Question 12 Deleted for CBSE Board 2024 Exams

Question 13 Deleted for CBSE Board 2024 Exams

Chapter 13 Class 12 Probability

Serial order wise

Last updated at April 16, 2024 by Teachoo

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) = (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