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Expert Answer
Let $S$ be the sample space that contains the possible sum of numbers when two dice are rolled, then: $S=\{2,3,4,5,6,7,8,9,10,11,12\}$ Since a pair of dice is rolled, therefore the total number of samples is $36$. Let $x$ denote the sums in the sample space and let $p$ be their probabilities then:$x$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ | $9$ | $10$ | $11$ | $12$ |
$p$ | $\dfrac{1}{36}$ | $\dfrac{2}{36}$ | $\dfrac{3}{36}$ | $\dfrac{4}{36}$ | $\dfrac{5}{36}$ | $\dfrac{6}{36}$ | $\dfrac{5}{36}$ | $\dfrac{4}{36}$ | $\dfrac{3}{36}$ | $\dfrac{2}{36}$ | $\dfrac{1}{36}$ |
$xp$ | $\dfrac{2}{36}$ | $\dfrac{6}{36}$ | $\dfrac{12}{36}$ | $\dfrac{20}{36}$ | $\dfrac{30}{36}$ | $\dfrac{42}{36}$ | $\dfrac{40}{36}$ | $\dfrac{36}{36}$ | $\dfrac{30}{36}$ | $\dfrac{22}{36}$ | $\dfrac{12}{36}$ |
Example 1
Harry rolls a fair die. Let $X$ be the event that the multiple of two occurs. Find the probability of $X$.Solution
Let $S$ be the sample space, then the possible outcomes are: $S=\{1,2,3,4,5,6\}$ Number of sample points in the sample space $n(S)=6$ The required outcomes are $2,4,6$. Now, $P(X)=\dfrac{\text{Number of favorable outcomes}}{\text{Total Outcomes}}$ $P(X)=\dfrac{3}{6}$ $P(X)=\dfrac{1}{2}$ Hence, the probability of Harry getting a multiple of $2$ is $\dfrac{1}{2}$.Example 2
A fair die is rolled $300$ times and there are $20$ chances of getting a $4$. Find the probability of getting a $4$.Solution
Let $X$ be the probability of getting a $4$ then: $P(X)=\dfrac{20}{300}$ $=\dfrac{2}{30}$ $P(X)=\dfrac{1}{15}$Previous Question < > Next Question
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