Question

1. use the bar - chart showing the number of pieces of each type of candy to determine the given probability.

a. you draw one piece of candy. what is the probability that you dont choose a grape candy?
b. you draw one piece of candy. what is the probability that you choose a lime or an orange candy?
c. you will draw three pieces of candy with replacement. what is the probability that you draw a cherry candy and then an apple or a lime candy?
d. you draw two pieces of candy without replacement. what is the probability that you draw a lime or an orange candy first and then a non - apple candy?

1. for this problem, assume that puppies are equally probable to be male or female. likewise for kittens.

a. our dog layla had two puppies. the older puppy is female. what is the probability that both puppies are female? (explain your answer)
b. our cat ariel had two kittens. at least one of them is male. what is the probability that both kittens are males? (hint: the answer is not 50%).

Explanation:

1a. Probability of not choosing grape candy

Let the number of apple candies be $a$, cherry candies be $c$, grape candies be $g$, lime candies be $l$ and orange candies be $o$. First, find the total number of candies $T=a + c+g + l+o$.
The probability of choosing a grape - flavored candy is $P(g)=\frac{g}{T}$.
The probability of not choosing a grape - flavored candy is $P(\text{not }g)=1 - P(g)=1-\frac{g}{T}=\frac{T - g}{T}$.

1b. Probability of choosing lime or orange candy

The number of lime and orange candies is $n=l + o$.
The total number of candies is $T$.
The probability of choosing a lime or orange candy is $P(l\cup o)=\frac{l + o}{T}$.

1c. Probability of drawing a cherry candy and then an apple or lime candy (with replacement)

The probability of drawing a cherry candy is $P(c)=\frac{c}{T}$.
The probability of drawing an apple or lime candy is $P(a\cup l)=\frac{a + l}{T}$.
Since the draws are with replacement, the two events are independent.
By the multiplication rule for independent events $P = P(c)\times P(a\cup l)=\frac{c(a + l)}{T^{2}}$.

1d. Probability of drawing a lime or orange candy first and then a non - apple candy (without replacement)

The number of lime or orange candies is $n_1=l + o$.
The total number of candies is $T$. The probability of drawing a lime or orange candy first is $P_1=\frac{l + o}{T}$.
After drawing one candy, the number of remaining candies is $T - 1$. The number of non - apple candies is $T - a$.
The probability of drawing a non - apple candy second is $P_2=\frac{T - a}{T - 1}$.
By the multiplication rule for dependent events $P = P_1\times P_2=\frac{(l + o)(T - a)}{T(T - 1)}$.

2a. Probability that both puppies are female given the older one is female

The sample space for the genders of two puppies is $\{(M,M),(M,F),(F,M),(F,F)\}$.
Since the older puppy is female, the reduced sample space is $\{(F,M),(F,F)\}$.
The probability that both are female given the older one is female is $\frac{1}{2}$.

2b. Probability that both kittens are male given at least one is male

The sample space for the genders of two kittens is $\{(M,M),(M,F),(F,M),(F,F)\}$.
The event that at least one is male is $E=\{(M,M),(M,F),(F,M)\}$.
The event that both are male is $A=\{(M,M)\}$.
The probability $P(A|E)=\frac{n(A\cap E)}{n(E)}=\frac{1}{3}$.

Answer:

1a. $\frac{T - g}{T}$
1b. $\frac{l + o}{T}$
1c. $\frac{c(a + l)}{T^{2}}$
1d. $\frac{(l + o)(T - a)}{T(T - 1)}$
2a. $\frac{1}{2}$
2b. $\frac{1}{3}$