Math problem ( Probabilities ) ..10 points for the right answer :) ?

Question

"You are eating a bowl of nuts. 40% of the nuts are almonds and the rest are peanuts, 45% of the almonds are salted. 65% of the peanuts are salted. You reach in and grab a nut. The probability that it is a rotten salted almond is 5%. The probability that it is an unsalted almond that is not rotten is 8%. The probability that it is rotten given that it is a peanut is 45%. You eat a nut and notice it is not rotten. What is the probability that it is a peanut?

Answer 1

The probability that a nut is not a rotten almond is
40%*(45*95%%+8*55%%) = 18.86%

The probability that a nut is not a rotten peanut is
60%*55% = 33%

So, the probability that a nut is not rotten and peanut is

33/(33+18.6)*100% = 63.63%

Answer 2

Nice problem!

40% almonds and 60% peanuts; can be subdivided as
(noting 40% * 45% = 18%; 60% * 65% = 39%)
18% salted almonds, 22% unsalted almonds, 39% salted peanuts, 21% unsalted peanuts.
This can be further broken down as
18% salted almonds (5% rotten, 13% not rotten),
22% unsalted almonds (14% rotten, 8% not rotten),
39% salted peanuts, 21% unsalted peanuts (combined 27% rotten, 33% not rotten).

P(peanut | not rotten) = P(peanut and not rotten) / P(not rotten)
= 33% / (33% + 8% + 13%)
= 33% / 54%
= 11/18
= 61% to the nearest percent.

Answer 3

Very nice! Thumbs up on your problem.

Given:

A = Almond
P = Peanut
S = Salted
~S = Not salted
R = Rotten
~R = Not rotten

Find p(P | ~R).

p(A) = .40
p(P) = .60
p(A ∩ S) = .40 * .45 = .18
p(P ∩ S) = .60 * .65 = .39
p(R ∩ A ∩ S) = .05
p(~R ∩ A ∩ ~S) = .08

p(A ∩ S) = p(R ∩ A ∩ S) + p(~R ∩ A ∩ S)
p(~R ∩ A ∩ S) = p(A ∩ S) - p(R ∩ A ∩ S) = .18 - .05 = .13

p(~R ∩ A) = p(~R ∩ A ∩ S) + p(~R ∩ A ∩ ~S) = .08 + .13 = .21

p(R | P) = .45
p(~R | P) = 1 - .45 = .55

p(~R ∩ P) = p(~R | P) * p(P) = .55 * .60 = .33

p(~R) = p(~R ∩ A) + p(~R ∩ P) = .21 + .33 = .54

p(P | ~R) = p(~R ∩ P)/p(~R) = .33/.54 = 11/18 = .6111 = 61.11%

Answer 4

It's quite a tricky one but there are 4 different outcomes that there is a white chip picked last, so you need to work out the probability for each of the outcomes (rrrww, rrwrw, rrwrw, wrrrw and rwrrw) and then add the fraction together, the answer should come to 33/120

Answer 5

not rotten peanut: 100 - 45 = 65% of total 60% peanuts.
0.65 * 60% =
39%

Answer 6

40% of the nuts are almonds and the rest are peanuts

thats all that is needed the rest are bull crap

60%

Answer 7

P(A) = 0.40
P(P) = 0.60
P(A n S) = 0.40 * 0.45
=0.18
P(P n S) = 0.60 * 0.65
= 0.39
P(A n S n R) = 0.18 * 0.05
= 0.009
P(A n D n R) = 0.40 * 0.55 * 0.92
= 0.2024
P(A n R) = 0.009 + 0.2024
= 0.2114
P(R/P) = 0.45 => [P(P n R)]/ P(P) = 0.45
=> P(P n R) = 0.45 *0.60 = 0.27
P(T) = 1 - 0.27 -0.2114
= 0.5186
P(P/T) = [P(P n T)] / P(T)
= 0.33 / 0.5186
= 0.6363
= 63.6%

A- Almond
P- Peanut
S- Salted
D- Not salted
R- Rotten
T - Not rotton

Answer 8

33 %

Answer 9

simple the answer is 60%.......................