A bag contrains 3 red marbles and 4 blue . A marble is taken at random from the bag and replaced. Another marble is taken from the bag. Work out the probability that the two marbles taken from the back or the same color. Give your answer as a fraction

4x-90(3)=50%m

Step-by-step explanation:

m= marbles (take note of my user name

For each draw, the chances of getting a blue marble are 3 of 7; the chances of getting a red are 4 of 7.

The chances of drawing a blue on both draws is (3/7)^2 or 9/49, and the chances of getting red on both draws is (4/7)^2 or 16/49. But there are other possible outcomes which have to be considered to come to the correct answer.

There are 4 possible outcomes, as follows (although the chances of these outcomes are different):

Comb Chances

RR …….. 16/49

RB …….. 12/49

BR …….. 12/49

BB ……… 9/49

total …….. 49/49

So we can say that the chances of drawing red marbles on each draw are 16/49, and the chances of drawing blue ones on each draw are 9/49. But what about the chances of drawing either both red OR both blue?

I believe we can simply add the 2 probabilities together to come up with a probability of

Ken Warren, former Accountant at Various (1977-2016)

Answered February 6, 2020

Originally Answered: There is a bag filled with 3 blue and 4 red marbles. A marble is taken at random from the bag, the colour is noted and then it is replaced. Another marble is taken at random. What is the probability of getting 2 of the same colour?

Probability of two reds = 4/7 x 4/7 = 16/49 = 32.65%

Probability of two blues = 3/7 x 3/7 = 9/49

Probability of the same colour = 25 / 49 = 51.0 %

Step-by-step explanation:

the answer would be c:   ax + b = ax + c

step-by-step explanation:

i think it is c hope that