Question

a bag contain 7 ball ,4are red , and 3 are green and one ball is drawn random (1) what are the chance of getting a red ball ? (2) what are the chance of getting a green ball? please answer​

Answer 1

Answer:

try this method to your question

Step-by-step explanation:

Let’s first assume that the balls must be drawn in the exact order that you state: white, white, white, then a red, then a green. That gives you FIVE balls, including one of each color, as well as 3 that are white.

First, let’s draw three white balls in a row from the bag of 23. The probability of doing that is:  723∗622∗521  

Next, from the remaining 20, the probability of drawing one of the 7 red balls is  720  

Finally, from the remaining 19, the probability of drawing one of the 4 green balls is  419  

Now, since there are 5! possible ways that the FIVE balls can be rearranged, as well as 18! different permutations of the remaining 18 undrawn balls, the product of the above probabilities must be divided by the product of 5! and 18!

Answer 2

Answer:

Here you go.

Step-by-step explanation:

Total balls = 7

Red balls = 4

Green " = 3

therefore, PROBABILITY (getting a red ball) = total outcomes/ outcomes of event

= 7/4

similarly ,

P ( getting a green ball) = 3/7.

(the reqd. answer)

you're welcome :)