Answer
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Hint: To calculate the odds of occurrence of 7, we first calculate the total sample space and sample space of the given event. Odds of an event referred to the chance of occurrence to the chance of not occurrence of the event. Sample space is nothing but the number of possibilities of an event.
Complete step-by-step solution:
The total number of possibilities of the event are:
Total number of outcomes of two dices = $6 \times 6 = 36$
Each dice can show up six different numbers.
Let E be the event “Sum of the number on the two dice is at least 7”.
Therefore, Sample space of E =
\\[E = \left\{
\left( {1,6} \right)\left( {2,5} \right)\left( {3,4} \right)\left( {4,3} \right)\left( {5,2} \right)\left( {6,1} \right) \\
\left( {3,5} \right)\left( {4,4} \right)\left( {5,3} \right)\left( {6,2} \right) \\
\left( {6,3} \right)\left( {4,5} \right)\left( {5,4} \right)\left( {3,6} \right) \\
\left( {6,4} \right)\left( {5,5} \right)\left( {4,6} \right) \\
\left( {5,6} \right)\left( {6,5} \right) \\
\left( {6,6} \right) \\
\right\}\]
Now, N (E), represents total sample space,
From above N (E) = 21
Also, we can write sample space for (Not E) because we know the total sample space is 36.
N (Not E) = 36 – 21 = 15
$\therefore {\text{P}}\left( {\text{E}} \right) = \dfrac{{21}}{{36}}{\text{ and P}}\left( {{\text{not E}}} \right) = \dfrac{{15}}{{36}}{\text{ - - - }}\left( 1 \right)$
Hence, In favor of event E,
\[ \Rightarrow P\left( E \right):P\left( {{\text{not }}E} \right) \\
\Rightarrow \dfrac{{P\left( E \right)}}{{P\left( {{\text{not }}E} \right)}} \]
Putting the values of P (E) and P (Not E) from equation (1), we get
\[\Rightarrow \dfrac{{\left( {\dfrac{{21}}{{36}}} \right)}}{{\left( {\dfrac{{15}}{{36}}} \right)}} \\
\Rightarrow \dfrac{{21}}{{15}} \\
\Rightarrow \dfrac{7}{5} \]
Note: In order to solve this type of problems the key is to always keep the total sample space of dice in our mind and in order to find the probability of an event, just divide the sample space of that event with the total sample space calculated. The probability of an event is defined as the ratio of favorable outcomes to the total number of outcomes of an event.
Complete step-by-step solution:
The total number of possibilities of the event are:
Total number of outcomes of two dices = $6 \times 6 = 36$
Each dice can show up six different numbers.
Let E be the event “Sum of the number on the two dice is at least 7”.
Therefore, Sample space of E =
\\[E = \left\{
\left( {1,6} \right)\left( {2,5} \right)\left( {3,4} \right)\left( {4,3} \right)\left( {5,2} \right)\left( {6,1} \right) \\
\left( {3,5} \right)\left( {4,4} \right)\left( {5,3} \right)\left( {6,2} \right) \\
\left( {6,3} \right)\left( {4,5} \right)\left( {5,4} \right)\left( {3,6} \right) \\
\left( {6,4} \right)\left( {5,5} \right)\left( {4,6} \right) \\
\left( {5,6} \right)\left( {6,5} \right) \\
\left( {6,6} \right) \\
\right\}\]
Now, N (E), represents total sample space,
From above N (E) = 21
Also, we can write sample space for (Not E) because we know the total sample space is 36.
N (Not E) = 36 – 21 = 15
$\therefore {\text{P}}\left( {\text{E}} \right) = \dfrac{{21}}{{36}}{\text{ and P}}\left( {{\text{not E}}} \right) = \dfrac{{15}}{{36}}{\text{ - - - }}\left( 1 \right)$
Hence, In favor of event E,
\[ \Rightarrow P\left( E \right):P\left( {{\text{not }}E} \right) \\
\Rightarrow \dfrac{{P\left( E \right)}}{{P\left( {{\text{not }}E} \right)}} \]
Putting the values of P (E) and P (Not E) from equation (1), we get
\[\Rightarrow \dfrac{{\left( {\dfrac{{21}}{{36}}} \right)}}{{\left( {\dfrac{{15}}{{36}}} \right)}} \\
\Rightarrow \dfrac{{21}}{{15}} \\
\Rightarrow \dfrac{7}{5} \]
Note: In order to solve this type of problems the key is to always keep the total sample space of dice in our mind and in order to find the probability of an event, just divide the sample space of that event with the total sample space calculated. The probability of an event is defined as the ratio of favorable outcomes to the total number of outcomes of an event.
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