Question

20 pairs of shoes are there in a closet. Four shoes are selected at random. The probability that there is at least one pair is

• A. $\frac{40}{40}\times \frac{38}{39}\times \frac{36}{38}\times \frac{34}{37}$
• B. $1-\frac{40}{40}\times \frac{38}{39}\times \frac{36}{38}\times \frac{34}{37}$
• C. $\frac{1{-}^{20}{C}_4}{{}^{40}{C}_4}$
• D. $\frac{1{-}^{40}{C}_4}{{}^{20}{C}_4}$

Answer 1

The correct option is B $1-\frac{40}{40}\times \frac{38}{39}\times \frac{36}{38}\times \frac{34}{37}$

Since we have $20$ pairs of shoes i.e. $40$ shoes and we need the probability such that there is atleast one pair of shoes, we subtract the probability of getting no pair of shoes from $1$.
Probability of getting no pair = $\frac{40}{40}\times \frac{38}{39}\times \frac{36}{38}\times \frac{34}{37}$
The explanation for the above expression is that initially we are free to choose any shoe out of $40$, after that we cannot choose the other paired shoe remaining in the box of the one we have already chosen and so we are left with $38$ possibilities.
Proceed similarly to get the answer.
The answer to the question asked thus becomes $1-\frac{40}{40}\times \frac{38}{39}\times \frac{36}{38}\times \frac{34}{37}$