Category: Probability

Challenging Probability problem 6

The owner of a restaurant counts the number of banquets received by the restaurant at the end of each week. The probability that the restaurant receives at least two banquets in a randomly chosen week is 0.3. A week is considered busy if the restaurant receives at least two wedding banquets in the week.

Calculate the probability that in a period of 6 consecutive weeks, the 6th week is the second busy week, given that there is at most one busy week in the first four weeks.

Ans: 0.166

Solution

1

2

Challenging Probability problem 5

An unbiased die is thrown 6 times. Calculate the probabilities that the six scores obtained will

i) be 1, 2,3, 4, 5,6 in some order

ii) have a product which is an even number

iii) consists of exactly two 6’s and four odd numbers

Solution

dice prob

iii Alternative solution

6C2: 6 slots choose 2 slots to insert the two 6s. Remaining 4 slots automatically will be for the four odd numbers.
3^4: Each odd number has 3 choices. for 4 odd numbers will be 3^4
Probability = (6C2 x 3^4) / 6^6 = 5/192

 

Venn diagram, Inclusion and Exclusion Principle

Venn diagram, Inclusion and exclusion principle can be useful in solving permutations, combinations and probability problems.

It is especially useful in solving combinations where the cases are not mutually exclusive and hence addition principle cannot be applied.

Example 1

venn diagram pnc

venn diagram p2

Challenging Probability problem 2

Problem

Vegetarian club consists of 2 married couples and 8 singles. The club is to select a delegation of 4 members to participate in an overseas conference. Find the probability that the delegation contains exactly one married couple.

Answer: 8/45

Solution

Case 1: 1 couple + 1 of the other couple + 1 single

(2 couples choose 1 couple) X (2 choose 1 from the other couple) X (8 singles choose 1) = 32

Case 2: 1 couple and 2 singles

(2 couples choose 1 couple) X (8 singles choose 2) = 56

No restrictions  = 12 C 4 = 495

Probability = (32+56) / 495 = 8/45

Challenging Probability problem 1

Problem

A bag contains 10 orange-flavoured, 14 strawberry-flavoured and 16 cherry-flavoured sweets which are of identical shapes and sizes. Benny selects a sweet at random from the bag. If it is not cherry-flavoured, he replaces it and selects another sweet at random. He repeats the process until he obtain a cherry-flavoured sweet. Calculate the probability that

i) the first sweet selected is strawberry-flavoured and the fourth sweet is orange-flavoured;

ii) he selects an even number of sweets

Solution

prob