Question

Question 17
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 3 $=\frac{n\left(E_1\right)}{n\left(S\right)}=\frac{43}{500}=0.086$
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.