Respuesta :
To answer the three last questions we are going to use the binomial distribution.
A binomial distribution can be thought of as simply the probability of a SUCCESS or FAILURE outcome in an experiment or survey that is repeated multiple times. The binomial is a type of distribution that has two possible outcomes.
Since in this case, we don't know the number of M&M's we can think that the experiments are independent of each other, that is, that they are repeated.
The binomial function is given by
[tex]P(x)=\frac{n!}{(n-x)!x!}p^x(1-p)^{n-x}[/tex]where n is the number of times we are doing the experiment, x is the outcome we would like to have and p is the probability of succes in the experiment.
Compute the probability that three randomly selected peanut M&M's are all orange.
In this case we are going to take three M&M's, so n=3. Since we want them to be orange and we know that the probability of picking and orange one is 23% then p=.23 (it has to be express in decimal form). Finally we would like that all of them are orangem then x=3. Plugging this values in the binomial distribution we have
[tex]\begin{gathered} P(x=3)=\frac{3!}{(3-3)!3!}(.23)^3(1-.23)^{3-3} \\ =0.012 \end{gathered}[/tex]therefore, the probability that three randomly selected M&M's are all orange is 0.012 or 1.2%.
If you randomly select five peanut M&M's, compute that probability that none of them are orange.
In this case the number of selected M&M's is five so n=5, the proability of taking an orange one is p=.23 and, since we don't want an orange one x=0. Then
[tex]\begin{gathered} P(x=0)=\frac{5!}{(5-0)!0!}(.23)^0(1-.23)^{5-0} \\ =.271 \end{gathered}[/tex]therefore the probability that in five randomly selected M&M's none of them are orange is .271 or 27.1%.
If you randomly select five peanut M&M's, compute that probability that at least one of them is orange.
Since the problem says at least one we can find this probability in two ways:
0. We add the probabilities of choosing one and two and three and four and five.
,1. Since the probabilities have to add 1, we can substract to 1 the probability of not choosing any orange one.
The second way is more straightforward, so let's use that way, then
[tex]\begin{gathered} P(x\ge1)=1-P(x=0) \\ =1-0.271 \\ =0.729 \end{gathered}[/tex]therefore the probability that in five randomly selected M&M's at least one of them is orange is .729 or 72.9%.
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