Question

Two dice are tossed once. The probability of getting even number at the first die or a total of 8 is
A) \[\dfrac{1}{{36}}\]
B) \[\dfrac{3}{{36}}\]
C) \[\dfrac{{11}}{{36}}\]
D) \[\dfrac{5}{9}\]

Answer 1

Hint: First of all we have to calculate the total number of outcomes in sample space. Now two events A and B are considered for getting an even number and getting a total of 8 respectively. Now the sample space of A and B is written. Now the \[A\bigcap B \] space is written. The probability of A, B, \[A\bigcap B \] is calculated and at last the required probability is calculated as follows.

Complete step by step answer:
In a single unbiased die, we can get any one of the outcomes after thrown: 1, 2, 3, 4, 5 or 6
Hence the total number of exhaustive outcomes is 6.
Now the same type of another die is also thrown which probable outcome will be the same as before i.e 6 as mentioned above. We know that in a single throw of two dies, the total number of possible outcomes is (6 × 6) = 36.
Let be \[S\] the sample space. Then, two dice are tossed, the total number of sample space,
\[ \Rightarrow n\left( S \right) = 36\]
Let \[A\] be the event of getting an even number on the first die and \[B\] be the event of getting a total of 8.
Now the favorable outcomes of A = {(2,1),(2,2),..,(2,6),(4,1),(4,2),..,(4,6),(6,1),(6,2),..,(6,6)}
\[ \Rightarrow n\left( A \right) = 18\]
Similarly the favorable outcomes of B = {{2,6),(3,5),(4,4),(5,3),(6,2)}}
\[ \Rightarrow n\left( B \right) = 5\]
Hence, \[A\bigcap {B = \left\{ {\left( {2,6} \right),\left( {4,4} \right),\left( {6,2} \right)} \right\}} \]
\[ \Rightarrow n\left( {A\bigcap B } \right) = 3\]
Now we know that the probability formula of an event as,
Probability, \[ P\left( E \right) = \dfrac{{total{\text{ number of possible outcomes }}}}{{the{\text{ total number of outcomes}}}}\]
Therefore, probability of getting even number in first die,
\[ \Rightarrow P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}} = \dfrac{{18}}{{36}}\]
Again, probability of getting total number of 8,
\[ \Rightarrow P\left( B \right) = \dfrac{{n\left( B \right)}}{{n\left( S \right)}} = \dfrac{5}{{36}}\]
Again probability of getting \[A\bigcap B \] is,
\[ \Rightarrow P\left( {A\bigcap B } \right) = \dfrac{{n\left( {A\bigcap B } \right)}}{{n\left( S \right)}} = \dfrac{3}{{36}}\]
Now for calculating the required probability of getting even number at the first die or a total of 8 is,
\[ \Rightarrow P\left( {A\bigcup B } \right) = P\left( A \right) + P\left( B \right) - P\left( {A\bigcap B } \right)\]
Now, putting the value in the above equation we get ,
\[ \Rightarrow P\left( {A\bigcup B } \right) = \dfrac{{18}}{{36}} + \dfrac{5}{{36}} - \dfrac{3}{{36}}\]
\[ \Rightarrow P\left( {A\bigcup B } \right) = \dfrac{{20}}{{36}}\]
\[ \Rightarrow P\left( {A\bigcup B } \right) = \dfrac{5}{9}\]
Hence, the correct answer is option (D) \[\dfrac{5}{9}\].

Note:
For solving such a type of probability question we should know about the probability formula, concept exhaustive event, mutually exclusive event and all the probability formulas should be known to us.