Question

There are 3 red, 3 white and 3 green balls in a bag. One ball is drawn at random from a bag:
P is the event that the ball is red.
Q is the event that the ball is not green.
R is the event that the ball is red or white.
S is the sample space.
Which of the following options is correct?
(A) $n\left( S \right)=9,n\left( P \right)=3,n\left( Q \right)=6,n\left( R \right)=6$
(B) $n\left( S \right)=4,n\left( P \right)=3,n\left( Q \right)=6,n\left( R \right)=6$
(C) $n\left( S \right)=6,n\left( P \right)=3,n\left( Q \right)=6,n\left( R \right)=6$
(D) $n\left( S \right)=7,n\left( P \right)=3,n\left( Q \right)=6,n\left( R \right)=6$

Answer 1

Hint: We have 3 red balls,3 white balls, and 3 three green balls. Here,$n\left( P \right)$ is the number of red balls. $n\left( Q \right)$ is the number of balls that are not green. If the ball is not green then, it can be either red or white. We can get the number of non-green balls $n\left( Q \right)$ , after adding the number of red balls and white balls. $n\left( R \right)$ is the number of balls that are either red or white. We can get the number of balls that are either red or white $n\left( R \right)$ , after adding the number of red balls and white balls. $n\left( S \right)$ is the sample space and which can be obtained by adding all the number of red balls, white balls, and green balls. Now, solve it further and the value of $n\left( P \right)$ , $n\left( Q \right)$ , $n\left( R \right)$ , and $n\left( S \right)$ .

Complete step-by-step answer:
According to the question, it is given that we have a bag in which some balls are colored with red, some are colored with white and some are colored with green.
The number of red balls = 3 …………………………….(1)
The number of white balls = 3 …………………………..(2)
The number of green balls = 3 …………………………(3)
We have an event P. P is the event that the ball is red. We have to find the value of $n\left( P \right)$ that is the number of red balls.
From equation (1), we have the number of red balls.
The number of red balls, $n\left( P \right)$ = 3 …………………………………(4)
From equation (1) and equation (2), we have the number of red and white balls.
The number of balls that are either red or white = $3+3=6$ ………………………(5)
Now, we have an event Q. Q is the event that the ball is not green. For event Q, we have to select those balls which are not green. If the ball is not green then, it can be either red or white.
We have to find the value of $n\left( Q \right)$ that is the number of balls that are not green.
We can get the number of non-green balls after adding the number of red balls and white balls.
From equation (1) and equation (2), we have the number of red and white balls.
Adding equation (1) and equation (2), we get
The number of balls which are not green, $n\left( Q \right)$ = 6 ……………………..(6)
Now, we have an event R. R is the event that the ball is red or white. We have to find the value of $n\left( R \right)$ that is the number of balls that are either red or white.
From equation (5), we have the number of balls that are either red or white.
The number of balls that are either red or white, $n\left( R \right)$ = 6 ……………………(7)
We know that sample space is the set of all possible outcomes or results of an experiment.
Here, S is the sample space. In the whole experiment, we can have only three outcomes. That is the balls can be red, white, or green.
So, the sample space S is the summation of the balls that are red, white, or green.
From equation (1), equation (2), and equation (3), we have the number of red balls, white balls, and green balls respectively.
The Sample space S, $n\left( S \right)$ = 3 + 3 + 3 = 9 ………………………..(8)
From equation (4), equation (6), equation (7) and equation (8), we have
$n\left( P \right)=3$ , $n\left( Q \right)=6$ , $n\left( R \right)=6$ , and $n\left( S \right)=9$ .
Hence, option (A) is the correct one.

Note: We can also solve this question, by figuring out the options given. In every option, only the value of $n\left( S \right)$ is changing while the value of $n\left( P \right)$ , $n\left( Q \right)$ , and $n\left( R \right)$ is the same in every option. So, if we only find the value of $n\left( S \right)$ then we can easily get the correct option. $n\left( S \right)$ is the summation of all balls which is equal to 9. Therefore, option (A) is the correct one.