A bag contains orange flavoured candies only. Rahi takes out one candy without looking into the bag. What is the probability that she takes out –
(i)The orange flavoured candy?
(ii) a lemon flavoured candy?
Answer
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Hint: Since the bag contains only orange flavoured candies only, so the event of Rahi taking out an orange flavoured candy is equal to the number of candies given. So the probability will be one. Now since there is no lemon flavored candy in the bag so it is impossible for Rahi to take out a lemon flavoured candy.
Complete step-by-step answer:
Let the event of taking out a candy out of a bag be E. Then
(i) Since the bag contains orange flavoured candies only .Therefore the chances of the candy being taken out of an orange flavoured candy will be equal to the total number of candies. Now we know that, Probability=$\dfrac{{{\text{No}}{\text{. of favourable outcomes}}}}{{{\text{Total number of outcomes}}}} = \dfrac{{{\text{Number of orange candies being taken out}}}}{{{\text{Total number of candies}}}}$
Since, Number of candies taken out is equal to the total number of candies, then
$ \Rightarrow {\text{P}}\left( {\text{E}} \right) = \dfrac{{{\text{Total number of candies}}}}{{{\text{Total number of candies}}}} = 1$
Answer-Hence the Probability of Rahi taking out an orange flavoured candy is$1$.
(ii) Since there are no lemon flavoured candies in the bag. So the event of taking out a lemon flavoured candy is $ = 0$
So , Probability=$\dfrac{{{\text{No}}{\text{. of favourable outcomes}}}}{{{\text{Total number of outcomes}}}} = \dfrac{{{\text{Number of lemon candies being taken out}}}}{{{\text{Total number of candies}}}}$
Since the event is zero so the Probability will also be $0$ .
$ \Rightarrow {\text{P}}\left( {\text{E}} \right) = 0$
The probability of Rahi taking out a lemon flavoured candy is $0$ .
Note: You can also solve this question by assuming the total number of candies to be x($ = 100{\text{ or 50}}$ ) .When you put the values in the formula, you’ll get the same answer. As lemon flavored candy is not mentioned so it is an impossible even. Hence probability is 0.
Complete step-by-step answer:
Let the event of taking out a candy out of a bag be E. Then
(i) Since the bag contains orange flavoured candies only .Therefore the chances of the candy being taken out of an orange flavoured candy will be equal to the total number of candies. Now we know that, Probability=$\dfrac{{{\text{No}}{\text{. of favourable outcomes}}}}{{{\text{Total number of outcomes}}}} = \dfrac{{{\text{Number of orange candies being taken out}}}}{{{\text{Total number of candies}}}}$
Since, Number of candies taken out is equal to the total number of candies, then
$ \Rightarrow {\text{P}}\left( {\text{E}} \right) = \dfrac{{{\text{Total number of candies}}}}{{{\text{Total number of candies}}}} = 1$
Answer-Hence the Probability of Rahi taking out an orange flavoured candy is$1$.
(ii) Since there are no lemon flavoured candies in the bag. So the event of taking out a lemon flavoured candy is $ = 0$
So , Probability=$\dfrac{{{\text{No}}{\text{. of favourable outcomes}}}}{{{\text{Total number of outcomes}}}} = \dfrac{{{\text{Number of lemon candies being taken out}}}}{{{\text{Total number of candies}}}}$
Since the event is zero so the Probability will also be $0$ .
$ \Rightarrow {\text{P}}\left( {\text{E}} \right) = 0$
The probability of Rahi taking out a lemon flavoured candy is $0$ .
Note: You can also solve this question by assuming the total number of candies to be x($ = 100{\text{ or 50}}$ ) .When you put the values in the formula, you’ll get the same answer. As lemon flavored candy is not mentioned so it is an impossible even. Hence probability is 0.
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