okeydonkey
New Member
- Joined
- Dec 31, 2019
- Messages
- 16
- Gender
- Female
- HSC
- 2020
Hi everyone, I am getting stuck on the following question since I cannot get an answer like the solution.
The probability of a particular team winning a match is 0.6, the probability of the team losing is 0.3 and the probability of a draw is 0.1. Two points are awarded for a win, one for a draw and none for a loss. The team must play four matches. Find the probability that the team will gain exactly four points
Here is my working out
There are 3 cases for 4 points (2,2,0,0) (2,1,1,0) and (1,1,1,0)
Hence the probability will be
4C2*0.6^2*0.3^2 + 4*3*0.6*0.1^2*0.3 + 4*0.3*0.1^3 = 0.2172
However, the answer provided by the booklet is 0.2161
Could someone please help me point out my mistake and guide me through the question also?
Thank you for the help!
The probability of a particular team winning a match is 0.6, the probability of the team losing is 0.3 and the probability of a draw is 0.1. Two points are awarded for a win, one for a draw and none for a loss. The team must play four matches. Find the probability that the team will gain exactly four points
Here is my working out
There are 3 cases for 4 points (2,2,0,0) (2,1,1,0) and (1,1,1,0)
Hence the probability will be
4C2*0.6^2*0.3^2 + 4*3*0.6*0.1^2*0.3 + 4*0.3*0.1^3 = 0.2172
However, the answer provided by the booklet is 0.2161
Could someone please help me point out my mistake and guide me through the question also?
Thank you for the help!