Three dice are rolled simultaneously. The probability of getting the sum of more than 5.
$
A)\dfrac{5}{{54}} \\
B)\dfrac{{67}}{{72}} \\
C)\dfrac{{103}}{{108}} \\
D)\dfrac{{53}}{{54}} \\
$
Answer
611.7k+ views
Hint: Find total possibilities and subtract from possibilities of getting 5 or less than 5
First, we start by finding out the total possible outcomes when three dice are rolled simultaneously. Then we separately find the probability that the number that can be summed by the three dices are less than or equal to 5. Then we subtract it from the total probability, to get the probability of getting the sum of more than 5 with three dices.
Complete step by step solution:
First, we calculate the total outcomes when three dice are rolled.
So,
The possible outcomes of the 3 dice rolled simultaneously are ${6^3} = 6 \times 6 \times 6 = 216$. It is N
Since we have the total outcomes, now calculate the probability of sums less than or equal to 5.
The possible sums that are obtained with three dices which are below and equal to 5 are
\[\left( {1,{\text{ }}1,{\text{ }}3} \right),{\text{ }}\left( {1,{\text{ }}3,{\text{ }}1} \right),{\text{ }}\left( {3,{\text{ }}1,{\text{ }}1} \right),{\text{ }}\left( {2,{\text{ }}2,{\text{ }}1} \right),{\text{ }}\left( {2,{\text{ }}1,{\text{ }}2} \right),{\text{ }}\left( {1,{\text{ }}1,{\text{ }}1} \right),{\text{ }}\left( {1,{\text{ }}1,{\text{ }}2} \right),{\text{ }}\left( {1,{\text{ }}2,{\text{ }}1} \right){\text{,}}\left( {2,{\text{ }}1,{\text{ }}1} \right),\left( {2,1,2} \right){\text{ }}and{\text{ }}\left( {1,{\text{ }}2,{\text{ }}2} \right)\]
The outcomes are equal to 10
This can be called as finding the probability of getting a total of at most 5.
$P(X < = 5) = \dfrac{{10}}{{216}}$.
Since, this is not what we have been asked to find out, we need to go further.
So, the above probability should subtract this from the total probability.
Hence,
P (Getting sum greater than 5) = 1- P (Getting sum at most 5)
$
P(X > 5) = 1 - \dfrac{{10}}{{216}} \\
= \dfrac{{216 - 10}}{{216}} \\
= \dfrac{{206}}{{216}} \\
= \dfrac{{103}}{{108}} \\
$
So, the correct answer is Option C.
Note: Here, we have been asked from greater than 5, we should be clear about it and make sure that we do not include the probabilities of having the sum 5. Also, it's important to find out the correct total outcomes of events.
First, we start by finding out the total possible outcomes when three dice are rolled simultaneously. Then we separately find the probability that the number that can be summed by the three dices are less than or equal to 5. Then we subtract it from the total probability, to get the probability of getting the sum of more than 5 with three dices.
Complete step by step solution:
First, we calculate the total outcomes when three dice are rolled.
So,
The possible outcomes of the 3 dice rolled simultaneously are ${6^3} = 6 \times 6 \times 6 = 216$. It is N
Since we have the total outcomes, now calculate the probability of sums less than or equal to 5.
The possible sums that are obtained with three dices which are below and equal to 5 are
\[\left( {1,{\text{ }}1,{\text{ }}3} \right),{\text{ }}\left( {1,{\text{ }}3,{\text{ }}1} \right),{\text{ }}\left( {3,{\text{ }}1,{\text{ }}1} \right),{\text{ }}\left( {2,{\text{ }}2,{\text{ }}1} \right),{\text{ }}\left( {2,{\text{ }}1,{\text{ }}2} \right),{\text{ }}\left( {1,{\text{ }}1,{\text{ }}1} \right),{\text{ }}\left( {1,{\text{ }}1,{\text{ }}2} \right),{\text{ }}\left( {1,{\text{ }}2,{\text{ }}1} \right){\text{,}}\left( {2,{\text{ }}1,{\text{ }}1} \right),\left( {2,1,2} \right){\text{ }}and{\text{ }}\left( {1,{\text{ }}2,{\text{ }}2} \right)\]
The outcomes are equal to 10
This can be called as finding the probability of getting a total of at most 5.
$P(X < = 5) = \dfrac{{10}}{{216}}$.
Since, this is not what we have been asked to find out, we need to go further.
So, the above probability should subtract this from the total probability.
Hence,
P (Getting sum greater than 5) = 1- P (Getting sum at most 5)
$
P(X > 5) = 1 - \dfrac{{10}}{{216}} \\
= \dfrac{{216 - 10}}{{216}} \\
= \dfrac{{206}}{{216}} \\
= \dfrac{{103}}{{108}} \\
$
So, the correct answer is Option C.
Note: Here, we have been asked from greater than 5, we should be clear about it and make sure that we do not include the probabilities of having the sum 5. Also, it's important to find out the correct total outcomes of events.
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