Probability Part 2

Probability Part 2

Ex. In a through of a coin find the probability of getting a tail. Solution : In this case sample space, S = { H, T } , Event E = { T } Ex. An unbiased die is tossed. Find the probability of getting of getting a multiple of 2. Solution : Here Sample space S = { 1, 2, 3, 4, 5, 6 }, Event E = { 2, 4, 6 } multiple of 2 Ex. An unbiased die is tossed. Find the probability of getting a number less than or equal to 4. Solution : Here Sample space S = { 1, 2, 3, 4, 5, 6 }, Event E = { 1, 2, 3, 4 } number less than or equal to 4. Ex. Two coins are tossed. What is the probability of getting (a) At most one head ? Solution : n(S) = { (HH), (HT), (TH), (TT) } = 4 n(E) = { HT, TH, TT } = 3 at most one head (b) At most two heads ? Solution : n(S) = { (HH), (HT), (TH), (TT) } = 4 n(E) = { (HH), (HT), (TH), (TT) }= 4 Ex. What is the chance that a leap year selected randomly will will have 53 sundays ? Solution : A leap year has 366 days, out of which there are 52 weeks and 2 more days. 2 more days can be (Sunday, Monday) (Monday, Tuesday) (Tuesday, Wednesday) (Wednesday, Thrusday) (Thrusday, Friday) (Friday, Saturday) (Saturday, Sunday) = n(S) = 7 So, (Sunday, Monday) and (Saturday, Sunday) = n(E) = 2, therefore chances that a leap year selected randomly will will have 53 sundays: Ex. What is the chance that a normal year selected randomly will will have 53 sundays ? Solution : A normal year has 365 days, out of which there are 52 weeks and 1 more day So, extra day can be Sunday, Monday, Tuesday, Wednesday, Thrusday, Friday, Saturday So, n(S) = 7 , n (E) = 1 Ex. When two dice are thrown, what is the probability that (a) Sum of numbers appeared is less than equal to 4 Solution : E = { (1,1) (1,2) (1,3) (2,1) (2,2) (3,1) } n(E) = 6 and n(S) = 36 (b) Sum of numbers is a multiple of 4 Solution : E= { (1,3) (2,2) (2,6) (3,1) (3,5) (4,4) (5,3) (6,2) (6, 6) } n(E) = 9, n (S) = 36 (c) Numbers appeared are equal Solution : E = { (1,1) (2,2) (3,3) (4,4) (5,5) (6,6) } n(E) = 6, n(S) = 36