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Question: A bag contains \[4\] red, \[5\] black and \[6\] white balls. A ball is drawn from the bag at random....

A bag contains 44 red, 55 black and 66 white balls. A ball is drawn from the bag at random. Find the probability that the ball drawn is: red or white.

Explanation

Solution

The formula that is needed to find the probability is
P(R)=n(R)n(S)P(R) = \dfrac{{n(R)}}{{n(S)}} ,
where n(R)n(R) is no. of favorable outcome and n(S)n(S) is total no. of events in the sample space.
The probability of two disjoint events AA or BB is given by
P(AorB)=P(A)+P(B)P(AorB) = P(A) + P(B)

Complete step by step answer:
It is given that the bag contains 44 red, 55 black and 66 white balls.
Then the sample space is S=R,R,R,R,B,B,B,B,B,W,W,W,W,W,WS = \\{ R,R,R,R,B,B,B,B,B,W,W,W,W,W,W\\}
Therefore, the total number of balls in the bag is 4+5+6=154 + 5 + 6 = 15
That is, the total no. of event in the sample n(S)=15n(S) = 15
To find: Probability of getting a red or white ball.
Let RR be the event of getting a red ball, then the probability of getting a red ball is given by
P(R)=n(R)n(S)P(R) = \dfrac{{n(R)}}{{n(S)}} ,
where n(R)n(R) is no. of favorable outcome and n(S)n(S) is total no. of events in the sample space.
From the sample space we get, n(R)=4n(R) = 4
Therefore, P(R)=415P(R) = \dfrac{4}{15}
Let WW be the event of getting a white ball, then the probability of getting a white ball is given by
P(W)=n(W)n(S)P(W) = \dfrac{{n(W)}}{{n(S)}} ,
where n(W)n(W) is no. of favorable outcome and n(S)n(S) is total no. of events in the sample space.
Again, from the sample space we get, n(W)=6n(W) = 6
Therefore, P(W)=615P(W) = \dfrac{6}{{15}}
Let AA be the event of getting a red or white ball, then the probability of AA is given by
P(A)=P(R)+P(W)P(A) = P(R) + P(W)
Therefore, P(A)=415+615P(A) = \dfrac{4}{{15}} + \dfrac{6}{{15}}
Simplifying this we will get,
P(A)=(4+6)15\Rightarrow P(A) = \dfrac{{(4 + 6)}}{{15}}
P(A)=1015\Rightarrow P(A) = \dfrac{{10}}{{15}}
Thus, the probability of getting a red or white ball is 1015\dfrac{{10}}{{15}}

Note: In this problem both the events are disjoints that is event of getting red ball and event of getting white ball are disjoint event (i.e. There is no intersection between these two events) so we used the formula P(AorB)=P(A)+P(B)P(AorB) = P(A) + P(B) . If the events are not disjoint events, then we have to use the formula P(AorB)=P(A)+P(B)P(AandB)P(AorB) = P(A) + P(B) - P(AandB) where P(AandB)P(AandB) is intersection between the events AA and BB .