22. A bag contains 4 red and 8 blue marbles. A marble is drawn at random. What is the<br />probability of drawing.<br />(i) a red marble?<br />(ii) a blue marble<br />23. A bag contains 4 green, 6 black and 7 white balls. A ball is drawn at random. What is the<br />probability that it is either a green or a black ball?
Answers
1st question:
Given that,
- number of red marbles = 4
- number of blue marbles = 8
Total number of events = (no. of red marbles) + (no. of blue marbles) = 4 + 8 = 12
We know that,
P(E) = (Number of events in which favourable outcome is produced)/(Total number of possible outcomes)
P(drawing a red marble) = (No. of red marbles)/(Total no. of marbles)
= 4/12
= 1/3
P(drawing a blue marble) = (No. of blue marbles)/(Total no. of marbles)
= 8/12
= 2/3
2nd question:
Given that,
- Number of green balls = 4
- Number of black balls = 6
- Number of white balls = 7
First we have to individually find P(getting green ball), P(getting black ball).
Total no. of balls = 4 + 6 + 7 = 17
We know that, P(E) = (Number of events in which favourable outcome is produced)/(Total number of possible outcomes)
So,
P(getting green ball) = (No. of green balls)/(Total no. of balls)
= 4/17
And,
P(getting black ball) = (No. of black ball)/(Total number of balls)
= 6/17
Remember that, when the word "OR" is used, we have to do addition (+), it is known as rule of sum. And when the word "AND" is used, we have to do multiplication (*), it is known as rule of product.
So,
P(getting a green or a black ball) = P(getting a green ball) + P(getting a black ball)
= 4/17 + 6/17
= 10/17