Probabilty

A word which can also be used in the place of ‘Probability’ is ‘Chance’. We all are well aware of the word ‘chance’. We use it in our daily life. Whatever we do, whatever we observe, there is always a chance that this is going to happen or not. When we want to find out the value of this chance in a quantitative form, at that point of time we use ‘Probability’ at the place of ‘Chance’.

Playing cards probability problems based on a well-shuffled deck of 52 cards.

Basic concept on drawing a card:

In a pack or deck of 52 playing cards, they are divided into 4 suits of 13 cards each i.e.

Cards of Spades and clubs are black cards. Total black coloured cards=26

Cards of hearts and diamonds are red cards. Total Red coloured cards=26

When cards are chosen from a pack of cards: Here we use the concept of combination because we are choosing (selecting) cards from a whole pack of cards. The total possible outcomes are 52 if only one pack of cards is used.

Question 1: Two cards are picked simultaneously from a pack of cards. What is the probability that both the cards will be queen?

Solution: Favourable outcomes – 2 (out of 4)

Total possible outcomes – 2 (out of 52)

Probability = ⁴C₂/⁵²C₂ = [(4x3)/(1x2)]/[(52x51)/(1x2)] = (4x3)/(52x51) = 1/221

Question 2: Two cards are picked out one by one from a pack of cards with replacement. What is the probability that both the cards will be queen?

Solution: First pick out –

Favourable outcomes – 1 (out of 4)

Total possible outcomes – 1 (out of 52)

Probability = ⁴C₁/⁵²C₁ = 4/52 = 1/13

Second pick out –

Favorable outcome – 1 (out of 4)

Total possible outcomes – 1 (out of 52)

Probability = ⁴C₁/⁵²C₁ = 4/52 =1/13

Because both events are happening, so final probability = (1/13) x (1/13) = 1/169

Question 3: Two cards are picked out one by one from a pack of cards without replacement. What is the probability that both the cards will be queen?

Solution: First pick out –

Favourable outcomes – 1 (out of 4)

Total possible outcomes – 1 (out of 52)

Probability = ⁴C₁/⁵²C₁ = 4/52 = 1/13

Second pick out –

Favourable outcome – 1 (out of 3)

Total possible outcomes – 1 (out of 51)

Probability = ³C₁/⁵¹C₁ = 3/51 = 1/17

Because both events are happening, so final probability = (1/13) x (1/17) = 1/221

Team/Group/Committee based probability problems:

• When team/group/committee is made with some constraints: In these questions, we have to find out the probability of different possibilities and we add those probabilities because only one combination of team/group/committee will be formed at a time.

A committee of 3 members is to be made out of 3 men and 2 women.

 Question 4: What is the probability that the committee has at least one woman?

Solution: Favourable outcomes – [1(out of 2 women) and 2(out of 3 men)]

Or [2(out of 2 women) and 1(out of 3 men)]

Total possible outcomes – 3 (out of 5)

Probability = [(²C₁x³C₂) + (²C₂x³C₁)] / ⁵C₃ = [(2x3) + (1x3)] / 10 = (6+3)/10 = 9/10

Question 5:  What is the probability that he committee has at most one woman?

Solution: Favourable outcomes – [0(out of 2 women) and 3(out of 3 men)]

Or [1(out of 2 women) and 2(out of 3 men)]

Total possible outcomes – 3 (out of 5)

Probability = [(²C₀x³C₃) + (²C₁x³C₂)] / ⁵C₃ = (1+6)/10 = 7/10

So, keep in mind these basic concepts before moving ahead with probability questions.