Question 473880: Can someone help with this problem, I don't have a calculator:
43% of adults say cashews are their favorite kind of nut. You randomly select 12 adults and ask each to name his or her favorite nut. Find the probability that the number who say cashews are their favorite is a)exactly three, (b) at least four and (c) at most two. If convenient use technology to find the probabilities.
(a) P(3)= (round to the nearest thousandth as needed)
(b) P(x≥4)= (round to the nearest thousandth as needed)
(c) P(≤2)= (round to the nearest thousandth as needed)
Thank you!
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! All of these problems involve the binomial distribution. To verify your work that deals with the binomial distribution, check out this calculator.
a)
P(X = 3) = (12 C 3)*(0.43)^(3)*(1-0.43)^(12-3)
P(X = 3) = (12 C 3)*(0.43)^(3)*(0.57)^(12-3)
Note: 12 C 3 = (12!)/(3!(12-3)!) = 220
P(X = 3) = (220)*(0.43)^(3)*(0.57)^9
P(X = 3) = (220)*(0.079507)*(0.006351461955384057)
P(X = 3) = 0.11109685085107844837778
P(X = 3) = 0.111
So the answer is 0.111
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b)
To calculate P(X >= 4), it's easier to first calculate P(X < 4) and then subtract this probability from 1. So to find P(X < 4), add up the probabilities for X=0 up to X=3 (we're not including 4)
P(X = 0) = (12 C 0)*(0.43)^(0)*(1-0.43)^(12-0) = (12 C 0)*(0.43)^(0)*(0.57)^(12-0) = (1)*(0.43)^(0)*(0.57)^12 = (1)*(1)*(0.001176246293903439668001) = 0.001176246293903439668001
P(X = 1) = (12 C 1)*(0.43)^(1)*(1-0.43)^(12-1) = (12 C 1)*(0.43)^(1)*(0.57)^(12-1) = (12)*(0.43)^(1)*(0.57)^11 = (12)*(0.43)*(0.0020635899893042801193) = 0.010648124344810085415588
P(X = 2) = (12 C 2)*(0.43)^(2)*(1-0.43)^(12-2) = (12 C 2)*(0.43)^(2)*(0.57)^(12-2) = (66)*(0.43)^(2)*(0.57)^10 = (66)*(0.1849)*(0.00362033331456891249) = 0.044180375571010266680466
P(X = 3) = (12 ncr 3)*(0.43)^(3)*(1-0.43)^(12-3) = (12 ncr 3)*(0.43)^(3)*(0.57)^(12-3) = (220)*(0.43)^(3)*(0.57)^9 = (220)*(0.079507)*(0.006351461955384057) = 0.11109685085107844837778
So
P(X = 0)= 0.001176246293903439668001
P(X = 1)= 0.010648124344810085415588
P(X = 2)= 0.044180375571010266680466
P(X = 3)= 0.11109685085107844837778
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Now onto the main event...
P(X >= 4) = 1 - P(X < 4)
P(X >= 4) = 1- [P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)]
P(X >= 4) = 1- [ 0.001176246293903439668001 + 0.010648124344810085415588 + 0.044180375571010266680466 + 0.11109685085107844837778]
P(X >= 4) = 1-0.1671015970608
P(X >= 4) = 0.8328984029392
P(X >= 4) = 0.833
So the answer is 0.833
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c)
To calculate P(X <= 2), simply add up all the probabilities from X=0 to X=2. So you'll need to calculate the following first
P(X = 0) = (12 C 0)*(0.43)^(0)*(1-0.43)^(12-0) = (12 C 0)*(0.43)^(0)*(0.57)^(12-0) = (1)*(0.43)^(0)*(0.57)^12 = (1)*(1)*(0.001176246293903439668001) = 0.001176246293903439668001
P(X = 1) = (12 C 1)*(0.43)^(1)*(1-0.43)^(12-1) = (12 C 1)*(0.43)^(1)*(0.57)^(12-1) = (12)*(0.43)^(1)*(0.57)^11 = (12)*(0.43)*(0.0020635899893042801193) = 0.010648124344810085415588
P(X = 2) = (12 C 2)*(0.43)^(2)*(1-0.43)^(12-2) = (12 C 2)*(0.43)^(2)*(0.57)^(12-2) = (66)*(0.43)^(2)*(0.57)^10 = (66)*(0.1849)*(0.00362033331456891249) = 0.044180375571010266680466
So
P(X = 0)= 0.001176246293903439668001
P(X = 1)= 0.010648124344810085415588
P(X = 2)= 0.044180375571010266680466
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Now onto the main event...
P(X <= 2) = P(X = 0) + P(X = 1) + P(X = 2)
P(X <= 2) = 0.001176246293903439668001 + 0.010648124344810085415588 + 0.044180375571010266680466
P(X <= 2) = 0.056004746209723791764055
P(X <= 2) = 0.056
So the answer is 0.056
To verify other work that deals with the binomial distribution, check out this calculator.
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