Anyone know probability?

Question

The question reads...one out of four avocados bought from the store are rotten. I buy 10 avocados, if my recipe calls for 8, what is the probability that I will have enough avocados that arent rotten?

The answer says .5256, I just need to figure out how he got that.

Answer 1

modulo's answer is correct, but I'll explain it to in more detail if you're not familiar with the Bionmial Probability. You can use this probability if you only have two events occuring. In this case, you do (rotten or not rotten). The Bionmial Probability is defined as follows:

P(x) = n!/[(n-x)!x!]*p^x*q^[n-x], where:

n = sample size
x = # of successes
p = probability of success
q = probability of failure = (1-p)

In this problem,

n = 10 (# of avocados you buy = sample size)
x = depends
p = 0.25 (1 out of 4 is rotten), here we define a success as rotten
q = 1-p = 0.75, here we define failure as not rotten

In this problem, you want at least 8 out of your 10 avocados to be not rotten. This means that most, you want less than 2 avocados to be rotten. In terms of probability notation, you want:

P(x<=2), remember we defined 'x' as rotten. Therefore,

P(x<=2) = P(x=2) + P(x=1) + P(x=0). Now just calculate each of these probabilities separately using the equation above and add them together to get your total probability.

For 2 rotten avocados:
P(x=2) = (10!/[8!2!])*(0.25)^2*(0.75)^8 = 0.282 or 28.2%

For 1 rotten avocado:
P(x=1) = (10!/[9!1!])*(0.25)^1*(0.75)^9 = 0.188 or 18.8%

For 0 rotten avocados:
P(x=0) = (10!/[10!0!])*(0.25)^0*(0.75)^10 = 0.056 or 5.6%

Adding all these together, you get:

P(x<=2) = P(x=2)+P(x=1)+P(x=0) =

0.282+0.188+0.056 = 0.526 = 52.6%, which is your desired answer

Below is a website that has more problems with Binomial Probability.

----------

Hope this helps

Answer 2

It's been a while, so there's probably a more elegant way- but this is close to what I do:
There are three cases where you'll have enough- none bad, one bad, or two bad. Add the probabilities, and you'll be good.

Case 1: none bad.
P1=0.75^10 (all of the avocados are good)
Case 2: one bad
This case is slightly more complicated. The probability that a specific avocado is the only bad one is 0.75^9*0.25, but that only counts that specific one. Because you don't care if it's the first avocado or the last one that's bad, multiply by the number of possibly bad avocados- 10.
P2=0.75^9 * 0.25 *10
Case 3: two bad
This case is even more complicated. The probability that any specific pair of avocados is bad is 0.75^8 * 0.25^2. To get the number of possible pairs, you can use the choose function-
10 choose 2- it's 10!/(2! 8!). If you haven't heard of the choose function, there must be a more elegant way to get the probability.

I'll check back in case you edit your question again.

Answer 3

Binomial, you want 8 of the 10 to be good. That's the same as no more than 2 rotten.

k is number of rotten ones:
P(k<3) = P(k=0) + P(k=1) + P(k=2)

p=1/4

10C0p^0(1-p)^10 + 10C1p^1(1-p)^9 + 10C2p^2(1-p)^8

=.0563 + .1877 + .2816 = 0.5256

Yep!

Answer 4

4 C 1 all over 10 C 8

C is combinations.

Answer 5

good, though
1 out of 4 is
2 out of 8
is 2.5 out of 10
So it is 50/50 whether you will get 2 out of your 10 or 3, the 3 would only give you 7.

Answer 6

if i did it right 40% chance of having enough avacodos