Q1: A committee of 5 is to be formed from 6 men and 4 women. In how many ways can it be
formed, if at least two women are included in the committee?
This is a combination problem. We are given 6 men and 4 women. We are supposed to choose
5 members for the committee such that at least 2 women are selected.
Possible cases:
● 2 women + 3 men = 4C2*6C3 = 6*20 = 120
● 3 women + 2 men = 4C3*6C2 = 4*15 = 60
● 4 women + 1 man = 4C4*6C1 = 1*6 = 6
So, the answer is 120+60+6 = 186.
Q2: During the Second World War, gunners in RAF bombers returning from missions were
asked from which direction they were most frequently attacked by enemy fighter planes. The
majority of the answer was ‘from above and behind’.
Suppose you had been about to become a gunner on a bomber and had heard about this.
You might have thought ‘I’d better concentrate on the skies above and behind us’. Why might
it have been unwise to assume that this would be true of attacks on bombers in general?
Supervision bias
One thing that we need to observe here is that aircraft are returning with certain areas
damaged. However, they are returning back. Adding extra weight to the aircraft by strengthening
the vulnerable parts is useless because those parts will still be affected. Parts of the plane that
are less damaged need extra armor plating.
Q3: What is the probability of getting three-of-a-kind and a pair (i.e. three cards of one rank,
and two cards of another rank) in the same hand for a 5 card hand poker hand?
Solution:
Let’s divide the problem into two parts; parts a and b. In part, we have to choose 3 cards of the
same rank out of 52 cards. Number of ranks; 13. Humber of cards of the same ranks; 4. So the
number of possible ways is: 4C3*13C1. Similarly for the remaining two cards. We already have
taken 1 rank away so the number of possible ways are 4C2*12C1. In total, we will get
4C3*13C1 * 4C2*12C1 = 4*13*6*12 = 3744.
Q4: In how many ways can 18 girls and 14 boys be arranged in a line?
Since there is no condition on the arrangement. We can simply add the number of boys and
girls, which gives us 18 + 14 = 32 kids to arrange. So, in this case: 32! = 32 x 31 x 30 x ... x 3 x
2 x 1 = 263130836933693530167218012160000000. So, there are
263,130,836,933,693,530,167,218,012,160,000 ways to arrange 18 girls and 14 boys in a line.
Q5:CGiven 1234567890, count the number of permutations possible for this, such that no
element appears on its given location in this arrangement.
