Question

Two dice are thrown simultaneously $500$ times. Each time the sum of two numbers appearing on their tops is noted and recorded as given in the following table:

Sum	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	$10$	$11$	$12$
Frequency	$14$	$30$	$42$	$55$	$72$	$75$	$70$	$53$	$46$	$28$	$15$

If the dice are thrown once more, what is the probability of getting a sum. (i) $3$? (ii) more than $10$? (iii) less than or equal to $5$? (iv) between $8$ and $12$?

Answer 1

Step 1: Find the probability of getting a sum of $3$ when thrown:

Sum	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	$10$	$11$	$12$	Total
Frequency	$14$	$30$	$42$	$55$	$72$	$75$	$70$	$53$	$46$	$28$	$15$	$500$

Total number of observations taken into account is $n\left(S\right)=500$

Let $E$ be the event of getting the sum of $3$.

Number of event where the sum of $3$ is obtained is $=30$

$\Rightarrow n\left(E\right)\hspace{0.17em}=30$

We know that $P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}$

Substituting the values we get, $P\left(E\right)=\frac{30}{500}=0.06$

Step 2: Find the probability of getting a sum more than$10$ when thrown:

Let $E_1$ be the event of getting the sum of $morethan10$.

Number of event where the sum of $morethan10$ obtained is $=28+15=43$

$\Rightarrow n\left(E_1\right)\hspace{0.17em}=43$

We know that $P\left(E_1\right)=\frac{n\left(E_1\right)}{n\left(S\right)}$

Substituting the values we get, $P\left(E_1\right)=\frac{43}{500}=0.086$

Step 3: Find the probability of getting a sum less than or equal to $5$ when thrown:

Let $E_2$ be the event of getting the sum less than or equal to $5$.

Number of event where the sum less than or equal to $5$ obtained is $=14+30+42+55=141$

$\Rightarrow n\left(E_2\right)\hspace{0.17em}=141$

We know that $P\left(E_2\right)=\frac{n\left(E_2\right)}{n\left(S\right)}$

Substituting the values we get, $p\left(E_2\right)=\frac{141}{500}=0.282$.

Step 4: Find the probability of getting a sum between $8$ and $12$ when thrown:

Let $E_3$ be the event of getting the sum less than or equal to $5$.

Number of event where the sum less than or equal to $5$ obtained is $=70+53+46+28+15=127$

$\Rightarrow n\left(E_3\right)\hspace{0.17em}=127$

We know that $P\left(E_3\right)=\frac{n\left(E_3\right)}{n\left(S\right)}$

Substituting the values we get, $p\left(E_3\right)=\frac{127}{500}=0.254$.