In a lottery, a person chooses six different numbers at random from 1

1.	In a lottery, a person chooses six different numbers at random from 1 to 20. If these six numbers match with the six numbers already fixed by the lottery committee, he wins the prize. What is the probability of winning the prize in the game?
Answer» All numbers are different (given in question), this will be the same as picking r different objects from n objects which is ⁿc_r Here, n= 20 and r = 6(as we have to pick 6 different objects from 20 objects) Now we shall calculate the value of ²⁰C₆ = \(\frac{(20)!}{(20-6)!\times(6)!}\)as ⁿC_r = \(\frac{(n)!}{(n-r)!\times(r)!}\) i.e. ²⁰C₆ = 38760 Therefore, 38760 cases are possible, and in that only one them has prize, i.e. total no. of desired outcome is 1 As we know, Probability of occurrence of an event = \(\frac{Total\,no.of\,Desired\,outcomes}{Total\,no.of\,outcomes}\) Therefore, the probability of winning a prize is = \(\frac{1}{38760}\) Conclusion: Probability of winning the prize in the game is \(\frac{1}{38760}\)

In a lottery, a person chooses six different numbers at random from 1 to 20. If these six numbers match with the six numbers already fixed by the lottery committee, he wins the prize. What is the probability of winning the prize in the game?

Answer»

All numbers are different (given in question), this will be the same as picking r different objects from n objects which is ⁿc_r

Here, n= 20 and r = 6(as we have to pick 6 different objects from 20 objects)

Now we shall calculate the value of ²⁰C₆ = \(\frac{(20)!}{(20-6)!\times(6)!}\)as ⁿC_r = \(\frac{(n)!}{(n-r)!\times(r)!}\) i.e. ²⁰C₆ = 38760

Therefore, 38760 cases are possible, and in that only one them has prize, i.e. total no. of desired outcome is 1

As we know,

Probability of occurrence of an event

= \(\frac{Total\,no.of\,Desired\,outcomes}{Total\,no.of\,outcomes}\)

Therefore, the probability of winning a prize is

= \(\frac{1}{38760}\)

Conclusion: Probability of winning the prize in the game is \(\frac{1}{38760}\)