Probability 問題2 10分

First question:
This is a combination problem that, for choosing 5 questions randomly out of 10 questions, there are a total of 10C5 = 252 possible combinations
Then, for choosing 5 questions randomly out of 6 questions (since she can do 6 of all the questions), there are a total of 6C5 = 6 possible combinations
Therefore, the probability that the student can solve all 5 questions in the examination will be:
6/252 = 1/42
Second question:
First of all, the total possible no. of combinations when choosing 6 numbers randomly out of 40 numbers is 40C6 = 3838380.
In those combinations, when counting from 1,2,3,4,5,6, to 35,36,37,38,39,40, there are a total of 35 possible combinations.
Therefore, the required probability is:
35/3838380 = 1/109668

講真呢2條係crazy野

化到d數咁大...用maths係唔能夠解決，唔係３日３夜都做唔哂

第一條：6/10 x5/9 x 4/8 x 3/7 x 2/6

第二條：35/3838380
=1/109668

1. We may use multiple law of probability to find out the required probability.

The questions selected are without replacement. So,

P (the student can solve all five problems)
= 6/10 X 5/9 X 4/8 X 3/7 X 2/6
= 1/42


2. The no. of combination of winning numbers = 35

∵ The winning combinations can be: 1,2,3,4,5,6 . 2,3,4,5,6,7 . ... . 35,36,37,38,39,40.
Altogether 35 combinations.

All possible combinations for choosing 6 numbers between 1 and 40
= 40C6
= 3 838 380

So, the required probability
= 35 / 3 838 380
= 1 / 109 668