Monday, August 22, 2016

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 }

Probability problems
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

Probability problems
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.

Probability problems
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

Probability problems

(b) At most two heads ?

Solution : n(S) = { (HH), (HT), (TH), (TT) } = 4

n(E) = { (HH), (HT), (TH), (TT) }= 4

Probability problems
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:

Probability questions
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

Probability problems
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

Probability problems

(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

Probability problems

(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

Probability problems

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