
What is the probability of rolling a sum of \[9\] with two dice?
Answer
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Hint: In the question we are asked to calculate the probability of getting a sum of \[9\] when two dice are rolled. In a single throw of a dice, we cannot get a number more than \[6\]. So, we will make the pairs in such a way that the sum of two numbers is \[9\] and an individual number is not more than \[6\]. Then we will calculate the total number of favourable outcomes and divide by the total number of outcomes to get the required probability.
Complete step by step answer:
When a dice is rolled there is a chance of getting any number from \[1\] to \[6\]. So, there are total \[6\] possible outcomes in a single throw of a dice. So, when two dice are rolled simultaneously then, total possible outcomes will be; \[6 \times 6 = 36\].
Because each number of the first dice can pair with all the six numbers of the second dice. So, the total number of pairings will be \[6 \times 6 = 36\].
Now we have to get the pair of numbers in such a way that the sum of the numbers of an individual pair is \[9\].
Therefore, the favourable outcomes are: \[\left( {3,6} \right),\left( {4,5} \right),\left( {5,4} \right),\left( {6,3} \right)\].
So, total number of favourable outcomes \[ = 4\]
Total number of outcomes \[ = 36\]
\[\therefore \] the probability of getting a sum of \[9\] when two dice are thrown simultaneously \[ = \dfrac{4}{{36}}\]
Here we have used the concept that;
\[{\text{probability = }}\dfrac{{{\text{no}}{\text{. of favourable outcomes}}}}{{{\text{ total number of outcomes}}}}\]
\[\therefore \] the probability of getting a sum of \[9\] when two dice are thrown simultaneously \[ = \dfrac{1}{9}\]
Note:
One thing we have to keep in mind is that we cannot make a pair like \[\left( {2,7} \right)\] or \[\left( {1,8} \right)\]. This is because in a single throw we cannot get a number more than \[6\]. Another thing to note is that \[\left( {4,5} \right),\left( {5,4} \right)\] are two different outcomes and we cannot consider them as one. In the former case we got \[4\] on the first dice and \[5\] on the second dice. Whereas, in the latter we got \[5\] on the first dice and \[4\] on the second dice.
Complete step by step answer:
When a dice is rolled there is a chance of getting any number from \[1\] to \[6\]. So, there are total \[6\] possible outcomes in a single throw of a dice. So, when two dice are rolled simultaneously then, total possible outcomes will be; \[6 \times 6 = 36\].
Because each number of the first dice can pair with all the six numbers of the second dice. So, the total number of pairings will be \[6 \times 6 = 36\].
Now we have to get the pair of numbers in such a way that the sum of the numbers of an individual pair is \[9\].
Therefore, the favourable outcomes are: \[\left( {3,6} \right),\left( {4,5} \right),\left( {5,4} \right),\left( {6,3} \right)\].
So, total number of favourable outcomes \[ = 4\]
Total number of outcomes \[ = 36\]
\[\therefore \] the probability of getting a sum of \[9\] when two dice are thrown simultaneously \[ = \dfrac{4}{{36}}\]
Here we have used the concept that;
\[{\text{probability = }}\dfrac{{{\text{no}}{\text{. of favourable outcomes}}}}{{{\text{ total number of outcomes}}}}\]
\[\therefore \] the probability of getting a sum of \[9\] when two dice are thrown simultaneously \[ = \dfrac{1}{9}\]
Note:
One thing we have to keep in mind is that we cannot make a pair like \[\left( {2,7} \right)\] or \[\left( {1,8} \right)\]. This is because in a single throw we cannot get a number more than \[6\]. Another thing to note is that \[\left( {4,5} \right),\left( {5,4} \right)\] are two different outcomes and we cannot consider them as one. In the former case we got \[4\] on the first dice and \[5\] on the second dice. Whereas, in the latter we got \[5\] on the first dice and \[4\] on the second dice.
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