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

$\begin{array}{cc}\text{Sum}& \text{Frequency}\\ 2& 14\\ 3& 30\\ 4& 42\\ 5& 55\\ 6& 72\\ 7& 75\\ 8& 70\\ 9& 53\\ 10& 46\\ 11& 28\\ 12& 15\end{array}$

If the dice are thrown once more, then 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

The total number of times, two dice are thrown simultaneously, n(S) = 500.

(i) Number of times getting a sum 3, n(E)= 30
$\therefore$ Probability of getting a sum 3 $=\frac{n\left(E\right)}{n\left(S\right)}=\frac{30}{500}=\frac{3}{50}=0.06$
Hence, the probability of getting a sum 3 is 0.06.

(ii) Number of times getting a sum more than 10, $n\left(E_1\right)=28+15=43$
$\therefore$ Probability of getting a sum more than 10 $=\frac{n\left(E_1\right)}{n\left(S\right)}=\frac{43}{500}=0.086$
Hence, the probability of getting a sum more than 10 is 0.086.

(iii) Number of times getting a sum less than or equal to 5,
$n\left(E_2\right)=55+42+30+14=141$
$\therefore$ Probability of getting sum less than or equal to 5 $=\frac{n\left(E_2\right)}{n\left(S\right)}=\frac{141}{500}=0.282$
Hence, the probability of getting a sum less than or equal to 5 is 0.282.

(iv) Number of times getting a sum between 8 and 12,
$n\left(E_3\right)=53+46+28=127$
$\therefore$ Required probability $=\frac{n\left(E_3\right)}{n\left(S\right)}=\frac{127}{500}=0.254$
Hence, the probability of getting a sum between 8 and 12 is 0.254.