-
damilolaamele
- Junior | Next Rank: 30 Posts
- Posts: 11
- Joined: Sat Aug 11, 2012 1:59 pm
In a basketball contest, players must make 10 free throws. Assuming a player has 90% chance of making each of his shots, how likely is it that he will make all of his first 10 shots?
[spoiler]Ans: The probability of making all of his first 10 shots is given by
(9/10)* (9/10)* (9/10)* (9/10)* (9/10)* (9/10)* (9/10)* (9/10)* (9/10)* (9/10) = (9/10)^10 = 0.348 => 35%[/spoiler]
Please can someone explain how (9/10)^10 was broken down to 0.348 without using a calculator or having to do a messy multiplication?
Thanks!
Source: GMAT Math Tough Problems. I got document somewhere on the BTG website but I can't remember where exactly.
[spoiler]Ans: The probability of making all of his first 10 shots is given by
(9/10)* (9/10)* (9/10)* (9/10)* (9/10)* (9/10)* (9/10)* (9/10)* (9/10)* (9/10) = (9/10)^10 = 0.348 => 35%[/spoiler]
Please can someone explain how (9/10)^10 was broken down to 0.348 without using a calculator or having to do a messy multiplication?
Thanks!
Source: GMAT Math Tough Problems. I got document somewhere on the BTG website but I can't remember where exactly.













