Question

A bag of jellybeans contains 10 cherry flavored jellybeans, 10 green apple flavored jellybeans, 10 lemon flavored jellybeans, and 10 blue raspberry flavored jellybeans. If Jaime eats one jellybean at a time at random, what is the probability that he eats a cherry flavored jellybean first and a green apple flavored jellybean second?

A. 3/52

B. 1/16

C. 79/156

D. 5/78


Please help i’ll give brainliest answer

Answer 1

Answer:

d

Step-by-step explanation:

Answer 2

The probability that Jaime eats a cherry flavored jellybean first and a green apple flavored jellybean second, for the given condition, is given by: Option D: 5/78

How to calculate the probability of an event?

Suppose that there are finite elementary events in the sample space of the considered experiment, and all are equally likely.

Then, suppose we want to find the probability of an event E.

Then, its probability is given as

$P(E) = \dfrac{\text{Number of favorable cases}}{\text{Number of total cases}} = \dfrac{n(E)}{n(S)}$

where favorable cases are those elementary events who belong to E, and total cases are the size of the sample space.

What is chain rule in probability?

For two events A and B, by chain rule, we have:

$P(A \cap B) = P(B)P(A|B) = P(A)P(B|A)$

where P(A|B) is probability of occurrence of A given that B already occurred.

For this case, we're specified that:

• Bag consists of 40 flavored jellybean, in which there are 10 cherry flavored jellybeans, 10 green apple flavored jellybeans, 10 lemon flavored jellybeans, and 10 blue raspberry flavored jellybeans.
• Jaime eats jellybean one at a time at random.
• P(The first jellybean he eats is cheery flavored and the second jellybean he eats is green apple flavored) = To evaluate.

Let we take:

A = event that the first jellybean he chose is cherry flavored.

Then, as there are 10 ways to choose a cherry flavored jellybean, and 40 ways to choose a single jellybean randomly, so we get:

n(A) = 10, and n(choosing a random jellybean from a bag of 40 jellybean) = 40

Thus, we get: P(A) = 10/40 = 1/4 = 0.25

Now, let we take B = event that after eating first jellybean (first can be of any flavor), the second jellybean he eats is of green apple flavored.

For second random picking, there are only 39 jellybeans remaining.

P(B|A) = Probability that the second jellybean chosen is green apple flavored given that the first jellybean was cherry flavored.

Since if first jellybean was cherry flavored, so there are still 10 green apple flavored jellybean.

Thus, P(B|A) = 10/39

Using the chain rule, we get:

P(First jellybean is cherry flavored and second jellybean is green apple flavored) = $P(A \cap B) = P(A)P(B|A) = \dfrac{10}{40} \times \dfrac{10}{39} = \dfrac{10}{156} = \dfrac{5}{78}$

Thus, the probability that Jaime eats a cherry flavored jellybean first and a green apple flavored jellybean second, for the given condition, is given by: Option D: 5/78

Learn more about probability here:

brainly.com/question/1210781