This is driving me buggy.
You have a set of 50 fruit. 10 of these fruit are apples. What is the probability of getting an apple if you choose 8 fruit at random?
TIA
What is the probability of NOT getting an apple.
Meaning,
No apple on first try.
No apple on second try.
No apple on third try.
....
Thus, the probability for each one respectively are,
(40)/(50)
(39)/(49)*
(38)/(48)
...
(33)/(43)
Thus,
(40*39*38*...*33)/(50*49*...*43)=(40P8)/(50P8)=.1432
This is the probability of not getting an apple.
Thus, the probability of getting an apple is,
1-.1432=.8568
*)Because you now have 1 less of each fruit after your chose the first one.
Hello, pig!
There are: .C(50,8) .= .(50!)/(8!42!) .= .536,878,650 possible samples.You have a set of 50 fruit. .10 of these fruit are apples.
What is the probability of getting an apple if you choose 8 fruit at random?
I assume that "getting an apple" means getting at least one apple.
The opposite of "at least one apple" is "no apples".
To get no apples, we choose 8 of the other 40 fruit.
. . There are: .C(40,8) .= .(40!)/(8!32!) .= .76,904,685 ways
Hence, there are: .536,878,650 - 76,904,685 .= .459,973,965 ways to get some apples.
. . . . . . . . . . . . . . . . . . . . . . .459,973,965
Therefore: . P(some apples) .= .---------------
. . . . . . . . . . . . . . . . . . . . . . .536,878,650
. . I'll let you reduce/simplify the fraction . . .