Respuesta :
From the problem, 40% are red and so 60% are not red.
The formula for calculating the probability of exact number of candies :
[tex]P(r)=_nC_r(a)^r(b)^{n-r}[/tex]where n = number of sample
r = number of specific samples
a = percentage of one sample
b = percentage of the other sample
The problem is looking for the probability that at least 3 of the candies in a box are red.
This is also the same as the probability that at most 3 of the candies in a box are NOT red subtracted from 1.
[tex]\begin{gathered} P(red)+P(notred)=1 \\ P(red)=1-P(notred) \end{gathered}[/tex]So we need to find the probability of having at most 3 candies are NOT red.
From the problem, n = 10 (because there are 10 candies in a box)
r = 0, 1 and 2 (these are the number of candies that are NOT red)
a = 40% or 0.4 (percentage of red candy)
b = 60% or 0.6 (percentage of not red candy)
The working equation will be :
[tex]P(red\ge3)=1-\lbrack P(0)+P(1)+P(2)\rbrack[/tex]So we need to find the probabilities of r = 0, 1 and 2.
For r = 0
[tex]_{10}C_0(0.4)^0(0.6)^{10-0}=0.00605[/tex]As you can see, the exponent of 0.4 is 0 because we need a candy that is NOT red. So it will be 0.
For r = 1
[tex]_{10}C_1(0.4)^1(0.6)^{10-1}=0.04031[/tex]For r = 2
[tex]_{10}C_2(0.4)^2(0.6)^{10-2}=0.12093[/tex]Using the working equation from above.
[tex]\begin{gathered} P(red\ge3)=1-\lbrack0.00605+0.04031+0.12093\rbrack \\ =1-0.16729 \\ =0.83271 \end{gathered}[/tex]The answer is A. 0.8327
