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Hint: - Here, simply we use the formula of probability by finding favorable outcomes and total outcomes.
For\[\left( {\text{i}} \right)\]
The bag contains only lemon flavored candies, and nothing else.
There are no orange flavored candies in the bag. Hence there is no
Possibility of taking out an orange candy.
Therefore the probability of taking out an orange flavored candy${\text{ = 0}}$.
For\[\left( {{\text{ii}}} \right)\]
The bag contains only lemon flavored candies, and nothing else. Let ${\text{E}}$ be the event of taking out a lemon flavored candy from a bag.
Therefore (Number of favorable outcomes) $ = $(Total number of possible outcomes)
We know that Probability of an event $P(E) = \dfrac{{\left( {{\text{Number of favorable outcomes}}} \right)}}{{{\text{(Total number of possible outcomes)}}}}$
$\therefore P(E) = 1$
Therefore, the probability of taking out a lemon flavored candy from bag $ = 1$
Note:-Whenever we face such types of questions for solving probabilities, always find a favorable outcome and total number of outcomes to get the result.
For\[\left( {\text{i}} \right)\]
The bag contains only lemon flavored candies, and nothing else.
There are no orange flavored candies in the bag. Hence there is no
Possibility of taking out an orange candy.
Therefore the probability of taking out an orange flavored candy${\text{ = 0}}$.
For\[\left( {{\text{ii}}} \right)\]
The bag contains only lemon flavored candies, and nothing else. Let ${\text{E}}$ be the event of taking out a lemon flavored candy from a bag.
Therefore (Number of favorable outcomes) $ = $(Total number of possible outcomes)
We know that Probability of an event $P(E) = \dfrac{{\left( {{\text{Number of favorable outcomes}}} \right)}}{{{\text{(Total number of possible outcomes)}}}}$
$\therefore P(E) = 1$
Therefore, the probability of taking out a lemon flavored candy from bag $ = 1$
Note:-Whenever we face such types of questions for solving probabilities, always find a favorable outcome and total number of outcomes to get the result.
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