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How do you calculate the probabilities, when you throw two dice, of obtaining (a) $a 5$, and (b) $an 11$?

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Last updated date: 04th Mar 2024
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IVSAT 2024
Answer
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Hint: Here the given question is based on the concept of probability. We have to find the probability of possibility of the dice , by substituting the given data to the definition of probability and on further simplification we get the required probability of possibility to throw off the dice.

Complete step-by-step solution:
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. Probability can range in from 0 to 1, where 0 means the event to be an impossible one and 1 indicates a certain event.
The probability formula is defined as the possibility of an event to happen is equal to the ratio of the number of favourable outcomes and the total number of outcomes.
\[\text{Probability of event to happen P}\left( E \right) = \dfrac{{\text{Number of favourable outcomes}}}{{\text{Total Number of outcomes}}}\]
The total number of possibilities of throwing two dices is \[6 \times 6 = 36\]
(a) a 5
When we throw two dice, the number 5 is obtained. The number of possibilities of obtaining 5 is
$(1, 4) (2, 3) (3,2) (4, 1)$.
There are 4 possibilities of obtaining the number 5.
The probability is \[P(A) = \dfrac{4}{{36}}\]
On simplifying we have
\[ \Rightarrow P(A) = \dfrac{1}{9}\]

(b) an 11
When we throw two dice, the number 11 is obtained. The number of possibilities of obtaining 11 is
$(6, 5) (5, 6)$.
There are 2 possibilities of obtaining the number 11.
The probability is \[P(B) = \dfrac{2}{{36}}\]
On simplifying we have
\[ \Rightarrow P(B) = \dfrac{1}{{18}}\]

Note: The probability is a number of possible values. Candidate must know we have to use the permutation concept or combination concept to solve the given problem because it is the first and main thing to solve the problem. Here we arrange the things in the possible ways so we are using the concept permutation.
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