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Hint: To find the probability involving the case when an unbiased die is thrown, we first find the total number of outcomes possible. Then we find the desired number of outcomes for the particular question and then use the definition of probability to find the answer.

Complete step-by-step solution:

First, we start with the meaning of the term probability. Probability helps us to find out the likelihood of an event to occur. For example, if in a weather forecast, it is said that there is 80% chance for the day to be rainy, this means that the probability for it to rain on that particular day is 0.8 (basically, we convert percentage to decimal by dividing by 100). Another example is that of a coin toss. The total number of outcomes in this case is 2 (heads and tails). To get the desired outcome (say heads), the probability is 0.5 since it is equally likely to be heads or tails.

Now, coming to the above question. First, we find out the total number of outcomes. The number of outcomes would be six (that is- 1,2,3,4,5 and 6). Now, the desired number of outcomes are the numbers less than 5. These include – 1,2,3 and 4 (that is, there are a total four outcomes). Now, to find probability,

Probability = $\dfrac{\text{Desired number of outcomes}}{\text{Total number of outcomes}}$

Probability = $\dfrac{4}{6}$

Probability = $\dfrac{2}{3}$

Hence, the required probability is $\dfrac{2}{3}$.

Note: An alternative to solve the problem is to use the following equation below-

Probability (number greater than equal to 5) + Probability (number less than 5) = 1 -- (1)

Now, we need to find the probability for the number to be less than 5. Thus, we can calculate the probability of the number greater than equal to 5. In this case, also the number of outcomes is 6. The desired number of outcomes is 2 (that is 5,6). Thus, calculating the probability is-

Probability = $\dfrac{\text{Desired number of outcomes}}{\text{Total number of outcomes}}$

Probability = $\dfrac{2}{6}$

Probability = $\dfrac{1}{3}$

Now, putting in expression (1), we get,

$\dfrac{1}{3}$ + Probability (number less than 5) = 1

Probability (number less than 5) = $\dfrac{2}{3}$

Hence, we can get back the same required probability by this method also.

Complete step-by-step solution:

First, we start with the meaning of the term probability. Probability helps us to find out the likelihood of an event to occur. For example, if in a weather forecast, it is said that there is 80% chance for the day to be rainy, this means that the probability for it to rain on that particular day is 0.8 (basically, we convert percentage to decimal by dividing by 100). Another example is that of a coin toss. The total number of outcomes in this case is 2 (heads and tails). To get the desired outcome (say heads), the probability is 0.5 since it is equally likely to be heads or tails.

Now, coming to the above question. First, we find out the total number of outcomes. The number of outcomes would be six (that is- 1,2,3,4,5 and 6). Now, the desired number of outcomes are the numbers less than 5. These include – 1,2,3 and 4 (that is, there are a total four outcomes). Now, to find probability,

Probability = $\dfrac{\text{Desired number of outcomes}}{\text{Total number of outcomes}}$

Probability = $\dfrac{4}{6}$

Probability = $\dfrac{2}{3}$

Hence, the required probability is $\dfrac{2}{3}$.

Note: An alternative to solve the problem is to use the following equation below-

Probability (number greater than equal to 5) + Probability (number less than 5) = 1 -- (1)

Now, we need to find the probability for the number to be less than 5. Thus, we can calculate the probability of the number greater than equal to 5. In this case, also the number of outcomes is 6. The desired number of outcomes is 2 (that is 5,6). Thus, calculating the probability is-

Probability = $\dfrac{\text{Desired number of outcomes}}{\text{Total number of outcomes}}$

Probability = $\dfrac{2}{6}$

Probability = $\dfrac{1}{3}$

Now, putting in expression (1), we get,

$\dfrac{1}{3}$ + Probability (number less than 5) = 1

Probability (number less than 5) = $\dfrac{2}{3}$

Hence, we can get back the same required probability by this method also.

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