Question

A bag contains eight yellow marbles nine Green marbles three purple marbles five red marbles to marbles are chosen from the bag what expression would give the probability that one marble is yellow and others read the expression that would best represent the solution

Answer 1

Answer:

P(Y and R) = P(Y)*P(R) + P(R)*P(Y)

P(Y and R) = 16/125 = 0.128 = 12.8%

Step-by-step explanation:

There are 8 Yellow marbles in the bag

There are 9 Green marbles in the bag

There are 3 Purple marbles in the bag

There are 5 Red marbles in the bag

The total number of marbles in the bag are

Total marbles = 8 + 9 + 3 + 5 = 25

We want to find the probability of selecting two marbles that is one Yellow marble and one Red marble from the bag.

The probability of selecting a Yellow marble is given by

P(Y) = number of Yellow marbles/total number of marbles

P(Y) = 8/25

The probability of selecting a Red marble is given by

P(Y) = number of Red marbles/total number of marbles

P(Y) = 5/25

P(Y) = 1/5

It is possible that the first marble selected is Yellow and the second is Red, and it is also possible that first marble selected is Red and the second is Yellow.

P(Y and R) = P(Y)*P(R) + P(R)*P(Y)

P(Y and R) = (8/25)*(1/5) + (1/5)*(8/25)

P(Y and R) = 16/125

P(Y and R) = 0.128

P(Y and R) = 12.8%

Answer 2

Answer:

probability of selecting one yellow and one red = 2/15

Step-by-step explanation:

We are told there are;

8 yellow marbles

9 green marbles

3 purple Marbles

5 red Marbles

Since two Marbles are selected,

Number of ways of selecting one yellow and one red is:

C(8,1) x C(5,1) = 8!/(1!(8 - 1)!) x 5!/(1!(5 - 1)!)

This gives 40

Now, the total number of Marbles in the question will be;

8 + 9 + 3 + 5 = 25 Marbles

Thus, number of ways to select any two Marbles from the total is;

C(25,2) = 25!/(2!(25 - 2)!) = 300

Thus; probability of selecting one yellow and one red = 40/300 = 2/15