Probability Question about Affording Loaves

[Byeongok]

A bakery sells 3 types of bread: raisin at 2 dollars a loaf, sourdough at 3 dollars a loaf and white at 1.5 dollars a loaf. There are twice the number of white loaves than sourdough and equal numbers of sourdough and raisin. Lillian is in a hurry and randomly selects 2 loaves and takes them to the counter to pay.
Lillian only has 4 dollars in her purse.

Show suitable workings to find the probability that Lillian will be able to afford her 2 chosen loaves.

Lillian Purse = 4 dollars
2 Raisin = 4 dollars
2 Sourdough = 6 dollars
2 White bread = 3 dollars

1 Raisin = 1 Sourdough = 2 White
(She can’t afford 2 of sourdough)

Pr (Can’t afford 2 Loaves) = 1/4
Pr (Can afford 2 Loaves) = 3/4 b. State any assumptions you have made.

She has ¾ chance that she can afford 2 loaves of bread, and ¼ chance she can’t afford 2 loaves of bread.

EDIT: I DONT KNOW WHY THE FONTS ARE WEIRD FOR THE FIRST PART D:
EDIT2: FIXED

 

[Jameson]

Hi Year10Student,

Welcome to MHB! :)

If we assume that she must pick two of the same kind, then you are correct.

Here's a way to do it through probability properties. We know that \(\displaystyle P(W)+P(S)+P(R)=1\) because the sum of all possible probabilities must equal 1. We also know that $P(S)=P(R)$ as well as $2P(S)=P(W)$. Combining this into the first equation we get that $2P(S)+P(S)+P(S)=1$, which means \(\displaystyle 4P(S)=1 \implies P(S)=\frac{1}{4}, P(R)=\frac{1}{4}, P(W)=\frac{1}{2}\)

She can afford two raisins or two whites, so if we add $P(R)+P(W)$ we indeed get \(\displaystyle \frac{3}{4}\) and get \(\displaystyle \frac{1}{4}\) for the sourdough, which she can't afford.

To do the problem this way we have assumed two things:

1) She will pick 2 loaves of the same kind of bread.
2) Each loaf is equally likely.

 

[Byeongok]

Hi Year10Student,

Welcome to MHB! :)

If we assume that she must pick two of the same kind, then you are correct.

Here's a way to do it through probability properties. We know that \(\displaystyle P(W)+P(S)+P(R)=1\) because the sum of all possible probabilities must equal 1. We also know that $P(S)=P(R)$ as well as $2P(S)=P(W)$. Combining this into the first equation we get that $2P(S)+P(S)+P(S)=1$, which means \(\displaystyle 4P(S)=1 \implies P(S)=\frac{1}{4}, P(R)=\frac{1}{4}, P(W)=\frac{1}{2}\)

She can afford two raisins or two whites, so if we add $P(R)+P(W)$ we indeed get \(\displaystyle \frac{3}{4}\) and get \(\displaystyle \frac{1}{4}\) for the sourdough, which she can't afford.

To do the problem this way we have assumed two things:

1) She will pick 2 loaves of the same kind of bread.
2) Each loaf is equally likely.

Thanks for the reply!

but would it be possible to find out the probability of her being able to afford 2 loaves that aren't the same type?

For example, 1 sourdough and 1 white bread or 1 raisin and 1 white.

Its confusing!