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Raghav has a collection of balls of different colours. He has a total of $35$ balls in his basket out of which seven are black in colour and eight are yellow in colour. Out of remaining balls, some are white and the rest are red.
Based on the above, answer the following questions :
(a) If the probability of drawing a red ball at random from the basket is three times that of a white ball, then find the number of red balls in the basket.
(b) Find the probability of drawing a ball at random from the basket which is either a black or a white ball.
Based on the above, answer the following questions :
(a) If the probability of drawing a red ball at random from the basket is three times that of a white ball, then find the number of red balls in the basket.
(b) Find the probability of drawing a ball at random from the basket which is either a black or a white ball.
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Total number of balls = $35$
Number of black balls = $7$
Number of yellow balls = $8$
(a) Number of white and red balls = $35 - 15 = 20$ ($\frac{1}{2}$ Mark)
Let number of white balls = $x$, then number of red balls = $(20 - x)$ ($\frac{1}{2}$ Mark)
Since, P (a red ball) = $3 \times$ P (a white ball), ($\frac{1}{2}$ Mark)
$\therefore \frac{20 - x}{35} = 3 \times \frac{x}{35}$ (1 Mark)
$\Rightarrow x = 5$ ($\frac{1}{2}$ Mark)
Number of white balls = $5$
and number of red balls = $15$ ($\frac{1}{2}$ Mark)
(b) P (a black ball) = $\frac{7}{35}$ ($\frac{1}{2}$ Mark)
P (a white ball) = $\frac{5}{35}$ ($\frac{1}{2}$ Mark)
P (either a black or a white ball) = $\frac{7 + 5}{35} = \frac{12}{35}$ (1 Mark)
Number of black balls = $7$
Number of yellow balls = $8$
(a) Number of white and red balls = $35 - 15 = 20$ ($\frac{1}{2}$ Mark)
Let number of white balls = $x$, then number of red balls = $(20 - x)$ ($\frac{1}{2}$ Mark)
Since, P (a red ball) = $3 \times$ P (a white ball), ($\frac{1}{2}$ Mark)
$\therefore \frac{20 - x}{35} = 3 \times \frac{x}{35}$ (1 Mark)
$\Rightarrow x = 5$ ($\frac{1}{2}$ Mark)
Number of white balls = $5$
and number of red balls = $15$ ($\frac{1}{2}$ Mark)
(b) P (a black ball) = $\frac{7}{35}$ ($\frac{1}{2}$ Mark)
P (a white ball) = $\frac{5}{35}$ ($\frac{1}{2}$ Mark)
P (either a black or a white ball) = $\frac{7 + 5}{35} = \frac{12}{35}$ (1 Mark)