
A container has $6$ yellow, $4$red and $8$ blue balls. What is the probability of drawing (i) yellow ball (ii) blue ball.
Answer
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Hint: Probability is defined as the ratio of no. of outcomes in an exhaustive set of equally likely outcomes that produce a given event to the total no. of possible outcomes. Probability of happen an event $'A'$ is given by, $P(A) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}$
Where $n\left( A \right)$ is the number of favorable outcomes and $n\left( S \right)$ is the total number of possible outcomes.
Use this formula to find the probability of drawing a yellow ball and blue ball.
Complete step-by-step answer:
Given, a container has $6$ yellow, $4$ red and $8$ blue balls.
We have to find the probability of drawing (i) yellow ball (ii) blue ball.
Let us consider that all the balls are equally likely to be drawn.
$\therefore $Total number of possible outcomes = $n\left( S \right) = 6 + 4 + 8 = 18$
(i) Let $'A'$ denote the event ‘a yellow ball is drawn’.
$\therefore $The number of outcomes favorable to the event=$n\left( A \right) = 6$
The probability of an event is given by,$P(A) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}$
On substituting the values, we get-
$P(A) = \dfrac{6}{{18}}$
$ \Rightarrow P(A) = \dfrac{1}{3}$
Therefore, the probability of drawing a yellow ball is $\dfrac{1}{3}$.
(ii) Let $'B'$ denote the event ‘a blue ball is drawn’.
$\therefore $The number of outcomes favorable to the event=$n\left( B \right) = 8$
The probability of an event is given by, $P(B) = \dfrac{{n\left( B \right)}}{{n\left( S \right)}}$
On substituting the values, we get-
$P(B) = \dfrac{8}{{18}}$
$ \Rightarrow P(B) = \dfrac{4}{9}$
Therefore, the probability of drawing a blue ball is $\dfrac{4}{9}$.
Note: In the questions of probability, the value of probability lies between $0$ and $1$,because the numerator (number of outcomes favorable to the event) is always less than or equal to the denominator( the number of all possible outcomes). Probability $0$ indicates impossibility of the event and $1$ indicates certainty of the event.
Where $n\left( A \right)$ is the number of favorable outcomes and $n\left( S \right)$ is the total number of possible outcomes.
Use this formula to find the probability of drawing a yellow ball and blue ball.
Complete step-by-step answer:
Given, a container has $6$ yellow, $4$ red and $8$ blue balls.
We have to find the probability of drawing (i) yellow ball (ii) blue ball.
Let us consider that all the balls are equally likely to be drawn.
$\therefore $Total number of possible outcomes = $n\left( S \right) = 6 + 4 + 8 = 18$
(i) Let $'A'$ denote the event ‘a yellow ball is drawn’.
$\therefore $The number of outcomes favorable to the event=$n\left( A \right) = 6$
The probability of an event is given by,$P(A) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}$
On substituting the values, we get-
$P(A) = \dfrac{6}{{18}}$
$ \Rightarrow P(A) = \dfrac{1}{3}$
Therefore, the probability of drawing a yellow ball is $\dfrac{1}{3}$.
(ii) Let $'B'$ denote the event ‘a blue ball is drawn’.
$\therefore $The number of outcomes favorable to the event=$n\left( B \right) = 8$
The probability of an event is given by, $P(B) = \dfrac{{n\left( B \right)}}{{n\left( S \right)}}$
On substituting the values, we get-
$P(B) = \dfrac{8}{{18}}$
$ \Rightarrow P(B) = \dfrac{4}{9}$
Therefore, the probability of drawing a blue ball is $\dfrac{4}{9}$.
Note: In the questions of probability, the value of probability lies between $0$ and $1$,because the numerator (number of outcomes favorable to the event) is always less than or equal to the denominator( the number of all possible outcomes). Probability $0$ indicates impossibility of the event and $1$ indicates certainty of the event.
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