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# A die is thrown. Find the probability that the number will appear to be greater than 3.

Last updated date: 14th Jul 2024
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Hint: We should know that the ratio of favourable outcomes to total number of outcomes is called probability. We know that from a die we get the outcomes are 1, 2, 3, 4, 5 and 6. So, now we have the total number of outcomes. Now we have to calculate the total number of favourable outcomes. Now we have to find a number of outcomes greater than 3. This will give the total number of favourable outcomes. Now we have to find the ratio of the number of favourable outcomes to the total number of outcomes. This will give us the probability that the number will appear to be greater than 3.

Before solving the question, we should know the definition of probability. The ratio of favourable outcomes to total number of outcomes is called probability. Now by using this definition, we can find the probability.
From the question, it was given that to find out the probability that the number will appear greater than 3. Now we have to find the total number of outcomes from a die and we have to find the total number of favourable outcomes.
From a die, the possible outcomes are $1,2,3,4,5,6$. So, the total number of outcomes are equal to 6. We know that the numbers greater than 3 and less than or equal to 6 are 4, 5, 6. So, the total numbers that are greater than 3 are equal to 3. So, the total number of favourable outcomes are equal to 3.
Let us assume the total number of favourable outcomes are equal to m, total number of outcomes are equal to n and the probability is equal to P(A). So, the ratio of m and n is equal to P(A).
\begin{align} & m=3....(1) \\ & n=6....(2) \\ & P(A)=\dfrac{m}{n}....(3) \\ \end{align}
Now we will substitute equation (1) and equation (2) in equation (3).
\begin{align} & \Rightarrow P(A)=\dfrac{3}{6} \\ & \Rightarrow P(A)=\dfrac{1}{2}....(4) \\ \end{align}
So, from equation (4), it is clear that the probability that the number will appear to be greater than 3 is equal to $\dfrac{1}{2}$.

Note: Students should be careful while reading the question. Some students may think that it was asked to find the probability of having a number that is greater than or equal to 3. Then we get the number of favourable outcomes is equal to 4. We already know that the number of outcomes is equal to 6.
Let us assume the total number of favourable outcomes are equal to m, total number of outcomes are equal to n and the probability is equal to P(A). So, the ratio of m and n is equal to P(A).
\begin{align} & m=4....(1) \\ & n=6....(2) \\ & P(A)=\dfrac{m}{n}....(3) \\ \end{align}
Now we will substitute equation (1) and equation (2) in equation (3).
\begin{align} & \Rightarrow P(A)=\dfrac{4}{6} \\ & \Rightarrow P(A)=\dfrac{2}{3}....(4) \\ \end{align}
So, from equation (4), it is clear that the probability that the number will appear to be greater than 3 is equal to $\dfrac{2}{3}$.