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What is the probability of getting a sum of $22$ or more when four dice are thrown?
(A) $\dfrac{1}{{456}}$
(B) $\dfrac{5}{{432}}$
(C) $\dfrac{9}{{756}}$
(D) $\dfrac{1}{{756}}$

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Last updated date: 22nd Jun 2024
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Answer
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Hint: Probability is the ratios of favourable outcomes and total outcomes of this event. Favorable outcomes will be arrangements where the sum is greater than or equal to $22$ . And total outcomes can be calculated by multiplying the number of choices of each place together. The maximum sum that can be obtained is $24$ . So find the number of arrangements to obtain the sum of $24,23{\text{ and 22}}$ for favorable outcomes.

Complete step-by-step answer:
Here in this problem, we are given that four dice are rolled and we need to find the probability of getting a sum of all four numbers on top of $22$ or more.
Before starting with the solution we need to understand the concept of probability first.
Probability is a measure of the likelihood of an event to occur. Many events cannot be predicted with total certainty. We can predict only the chance of an event to occur i.e. how likely they are to happen, using it. The value is expressed from zero to one. The meaning of probability is basically the extent to which something is likely to happen. This is the basic probability theory, which is also used in the probability distribution, where you will learn the possibility of outcomes for a random experiment. To find the probability of a single event to occur, first, we should know the total number of possible outcomes.
$ \Rightarrow $ Probability $ = \dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of outcomes}}}}$
When four dice are rolled together, with each having possibilities of $1,2,3,4,5{\text{ and }}6$.
So, when all four dices are rolled together then the total number of outcomes possible can be calculated by multiplying the number of choices on each of the places, i.e.
$ \Rightarrow $ Total possible outcomes $ = 6 \times 6 \times 6 \times 6 = {6^4} = 1296$
Now let’s try to figure out the arrangements where the sum is greater than or equal to $22$ .
$ \Rightarrow 6 + 6 + 6 + 6 = 24$ , this is the maximum possible sum that can be obtained here
$ \Rightarrow 6 + 5 + 6 + 6 = 23$ , this is the sum that can be obtained by $4$ ways, with different positions of the number $5$ to appear.
We should now write all the possible arrangements for sum $22$ using all four positions:
$ \Rightarrow 6 + 6 + 6 + 4 = 22$ , this arrangement is possible in $4$ ways with different positions of numbers $4$ to appear.
$ \Rightarrow 6 + 5 + 5 + 6 = 22$ , this arrangement is only possible in $6$ ways
Therefore, the total number of favourable outcomes is $ = 1 + 4 + 4 + 6 = 15$
Thus, by using the formula for probability, we get:
$ \Rightarrow $ Probability $ = \dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of outcomes}}}} = \dfrac{{15}}{{1296}} = \dfrac{5}{{432}}$
Hence, the option (B) is the correct answer.

Note: In probability, questions like this can be solved with careful arrangement of favourable outcomes. Be careful while finding them. For finding a number of arrangements of $6 + 5 + 5 + 6 = 22$ , we can use the arrangement of four things at four places but with two similar things. And this can be represented by: $\dfrac{{4!}}{{2! \times 2!}} = \dfrac{{1 \times 2 \times 3 \times 4}}{{1 \times 2 \times 1 \times 2}} = 6$ .