Question

Find the chance of throwing six with a single die at least once in five trials.

Answer 1

Hint: A normal die has six different faces. Determine the probability of getting 6 and not getting 6 in a single trial. Use the formula ${\text{Probability}} = \dfrac{{{\text{No}}{\text{. of favorable outcomes}}}}{{{\text{Total no}}{\text{. of outcomes}}}}$. Then first find the probability of not getting 6 in any of the trials by using the rule of probability of independent events. Finally apply the formula of probability of complement events i.e. $P + \bar P = 1$, to get the answer.

Complete step-by-step answer:
According to the question, a single die is thrown five times.
We know that a normal die has six different faces and they are 1, 2, 3, 4, 5 and 6. Thus the probability of throwing any one of them in a trial is determined by the probability formula:
$ \Rightarrow {\text{Probability}} = \dfrac{{{\text{No}}{\text{. of favorable outcomes}}}}{{{\text{Total no}}{\text{. of outcomes}}}}$
Here, the number of favorable outcomes is 1 and the total number of outcomes is 6. So we have:$ \Rightarrow {\text{Probability}} = \dfrac{1}{6}{\text{ }}.....{\text{(1)}}$
From the probability formula, we also know that:
$ \Rightarrow $Probability of not getting 6 in a trial = 1 – Probability of getting 6.
From equation (1), we have calculated that the probability of getting 6 in a trial is $\dfrac{1}{6}$. Using this, we’ll get:
$ \Rightarrow $Probability of not getting 6 in a trial $1 - \dfrac{1}{6} = \dfrac{5}{6}$
Now, we have to determine the probability of getting at least one 6 in five trials. Let this probability be $P$.
First we will determine the probability of not getting 6 in any of the trials. This will be denoted by $\bar P$ because it is the complement of getting at least one 6.
Also each trial is independent of the other, so the probabilities of each trial will be multiplied to get the overall probability. Thus we have:
$
   \Rightarrow \bar P = \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{5}{6} \\
   \Rightarrow \bar P = \dfrac{{3125}}{{7776}}
 $
From the probabilities of complement events, we know that:
$ \Rightarrow P + \bar P = 1$
Putting the value of $\bar P$ from above in this equation, we’ll get:
$
   \Rightarrow P + \dfrac{{3125}}{{7776}} = 1 \\
   \Rightarrow P = 1 - \dfrac{{3125}}{{7776}} \\
   \Rightarrow P = \dfrac{{4651}}{{7776}}
 $

Thus the probability of getting six at least once in five trials is $\dfrac{{4651}}{{7776}}$

Note: If two events $A$ and $B$ with probabilities of their occurrences $P\left( A \right)$ and $P\left( B \right)$, are independent to each other, then the probabilities of occurrence of both of them is given by the formula:
$ \Rightarrow P\left( {A{\text{ and }}B} \right) = P\left( A \right) \times P\left( B \right)$
If the two events are mutually exclusive instead, then the occurrence of any one of them is given by the formula:
$ \Rightarrow P\left( {A{\text{ or }}B} \right) = P\left( A \right) + P\left( B \right)$