$A$ bag contains a red ball,a blue ball and a yellow ball,all the balls being of the same size. Kritika takes out a ball from the bag without looking into it. What is the probability that she takes out the
$(i)$ yellow ball?
$(ii)$ red ball?
$(iii)$ blue ball?
$(i)$ yellow ball?
$(ii)$ red ball?
$(iii)$ blue ball?
(N/A) Kritika takes out a ball from the bag without looking into it. Since all balls are of the same size,it is equally likely that she takes out any one of them.
Let $Y$ be the event 'the ball taken out is yellow',$B$ be the event 'the ball taken out is blue',and $R$ be the event 'the ball taken out is red'.
The total number of possible outcomes $= 3$.
$(i)$ The number of outcomes favourable to the event $Y = 1$.
So,$P(Y) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} = \frac{1}{3}$.
$(ii)$ The number of outcomes favourable to the event $R = 1$.
So,$P(R) = \frac{1}{3}$.
$(iii)$ The number of outcomes favourable to the event $B = 1$.
So,$P(B) = \frac{1}{3}$.
Let $Y$ be the event 'the ball taken out is yellow',$B$ be the event 'the ball taken out is blue',and $R$ be the event 'the ball taken out is red'.
The total number of possible outcomes $= 3$.
$(i)$ The number of outcomes favourable to the event $Y = 1$.
So,$P(Y) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} = \frac{1}{3}$.
$(ii)$ The number of outcomes favourable to the event $R = 1$.
So,$P(R) = \frac{1}{3}$.
$(iii)$ The number of outcomes favourable to the event $B = 1$.
So,$P(B) = \frac{1}{3}$.
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| $3$ | $4$ | $5$ | $5$ | $6$ | $6$ | $9$ |
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$(iii)$ at least $6?$
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