Question

At a museum cafe you can get a pre-made boxed lunch with a sandwich, fruit, and drink for only $3.

The sandwiches are made with either turkey or ham.
The fruit is either an apple or an orange.
The drink is either bottled water or juice.
The number of boxes they make for every possible combination is the same. If you randomly choose one of the boxed lunches without knowing the contents, what is the probability you will get an orange and not get juice in your box?

Answer 1

Given that sandwiches are made with either turkey or ham.

Prob for turkey or ham = 1/2

The fruit can be either apple or orange. Hence p(apple) = p(orange) =12

Drinks can be either bottled water or juice

P(water) = P(juice) = 1/2

We find that sandwiches, fruits, and drinks are mutually independent of each other.

Hence the probability you will get an orange and not get juice in your box

Prob (you get an orange and bottled water)

= Prob (orange) *Prob (bottled water) (since the two are independent

= 1/2 (1/2)= 1/4

Hence answer is 0.25

Answer 2

The conditional probability of getting a specified combination of fruit and

drink is found by the product of each of the probabilities.

The probability of getting an orange as the fruit and not getting a juice as

the drink in the box is 0.25

Reasons:

The number of sandwiches having an apple as fruit = The number of

sandwiches having an orange as fruit

Therefore;

Probability of getting a sandwich with orange as the fruit, P(O) = $\dfrac{1}{2}$

Probability of getting a sandwich with apple as the fruit, P(O) = $\dfrac{1}{2}$

Similarly, we have;

The number of sandwiches that have water as the drink = The number of

sandwiches that have juice as the drink

Therefore;

Probability of getting a sandwich having water as the drink P(W) = $\dfrac{1}{2}$

Probability of getting a sandwich having juice as the drink P(J) = $\dfrac{1}{2}$

Probability of not getting juice  P(J') = 1 - P(J) =  Probability of getting water,

P(W) = $\dfrac{1}{2}$

The probability of getting an orange and not getting a juice is therefore;

P(O and J') = P(O) × (1 - P(J))

Which gives;

$P(O \ and \ J') = \dfrac{1}{2} \times \left(1 - \dfrac{1}{2} \right) = \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}$

The probability of getting an orange and not getting a juice in the box, P(O and J') = $\dfrac{1}{4}$ = 0.25

Learn more here:

https://brainly.com/question/12783373