- #1
domingoleung
- 7
- 1
- TL;DR Summary
- Cookies that might have hazelnuts or almonds.
Here's the problem:
A chef made 500 cookies randomly mixed with 1000 nuts including 600 almonds and 400 hazelnuts in which each nut is the same size. Suppose the number of pieces of nuts in a piece of cookie follows a Poisson distribution.
(a) Suppose cookies are randomly selected one-by-one with replacement, find the probability that the fifth chosen cookie will be the 1st cookie without any hazelnut?
(b) Suppose 12 cookies are packed into each box. What is the probability of getting a randomly selected box with at most two piecs of cookies without any nuts?
My Appraoch:
For (a), I have two two ways of doing so:
1.
λ hazelnuts = 400/500 = 0.8
P(cookies with 1 or more hazelnuts)4 * P(cookies without hazelnuts)
= (1- e-0.8 (0.8)0 )4 * e-0.8 (0.8)0
= 0.0413
2.
λ hazelnuts = 400/500 = 0.8
P(1st cookie without hazelnut) x 1/500
= e-0.8 (0.8)0 x 1/500
= 8.98 x 10-4
For (b),
First Approach:
λ nuts = 1000/500 = 2
P(no nuts)
= e-2 (2)0
= 0.1353
λ no nuts in 12 cookies = 12 x 0.1353 = 1.624
P(At most 2 cookies without any nuts)
= P(no cookies without nuts) + P(1 cookie without nuts) + P(2 cookies without nuts)
= e-1.624 (1.624)0 + e-1.624 (1.624)1 + e-1.624 (1.624)2 / 2!
= 0.7771
Second Approach:
λ nuts = 1000/500 = 2
P( 1 cookie without any nuts and 11 cookies with nuts) + P( 2 cookies with no nuts and 10 cookies with nuts ) + P(All cookies with nuts)
= (1 - e-2 (2)0)11 x e-2 (2)0 + 10C2 * (1 - e-2 (2)0)10 x (e-2 (2)0)2 + (1 - e-2 (2)0)12
= 0.3945
To me, the two approaches for each sub question just make sense to me. Please tell me if I did something wrong :/
A chef made 500 cookies randomly mixed with 1000 nuts including 600 almonds and 400 hazelnuts in which each nut is the same size. Suppose the number of pieces of nuts in a piece of cookie follows a Poisson distribution.
(a) Suppose cookies are randomly selected one-by-one with replacement, find the probability that the fifth chosen cookie will be the 1st cookie without any hazelnut?
(b) Suppose 12 cookies are packed into each box. What is the probability of getting a randomly selected box with at most two piecs of cookies without any nuts?
My Appraoch:
For (a), I have two two ways of doing so:
1.
λ hazelnuts = 400/500 = 0.8
P(cookies with 1 or more hazelnuts)4 * P(cookies without hazelnuts)
= (1- e-0.8 (0.8)0 )4 * e-0.8 (0.8)0
= 0.0413
2.
λ hazelnuts = 400/500 = 0.8
P(1st cookie without hazelnut) x 1/500
= e-0.8 (0.8)0 x 1/500
= 8.98 x 10-4
For (b),
First Approach:
λ nuts = 1000/500 = 2
P(no nuts)
= e-2 (2)0
= 0.1353
λ no nuts in 12 cookies = 12 x 0.1353 = 1.624
P(At most 2 cookies without any nuts)
= P(no cookies without nuts) + P(1 cookie without nuts) + P(2 cookies without nuts)
= e-1.624 (1.624)0 + e-1.624 (1.624)1 + e-1.624 (1.624)2 / 2!
= 0.7771
Second Approach:
λ nuts = 1000/500 = 2
P( 1 cookie without any nuts and 11 cookies with nuts) + P( 2 cookies with no nuts and 10 cookies with nuts ) + P(All cookies with nuts)
= (1 - e-2 (2)0)11 x e-2 (2)0 + 10C2 * (1 - e-2 (2)0)10 x (e-2 (2)0)2 + (1 - e-2 (2)0)12
= 0.3945
To me, the two approaches for each sub question just make sense to me. Please tell me if I did something wrong :/