Two dice are thrown $400$ times. Each time sum of two numbers appearing on their tops is noted as given in the following table:
Sum
$2$
$3$
$4$
$5$
$6$
$7$
$8$
$9$
$10$
$11$
$12$
Frequency
$14$
$20$
$32$
$45$
$62$
$65$
$60$
$43$
$36$
$18$
$5$
What is the probability of getting a sum (i) $5$ , (ii) more than $10$ , (iii) between $5 a n d 10$ ?

Open in App

Solution

Step 1: Find the probability of getting a sum of $5$ on the top:
Sum
Frequency
$2$
$14$
$3$
$20$
$4$
$32$
$5$
$45$
$6$
$62$
$7$
$65$
$8$
$60$
$9$
$43$
$10$
$36$
$11$
$18$
$12$
$5$
Total
$400$
Let $E$ be the event of getting a sum of $5$ on the top of two dice.
From the frequency distribution table, $n (E) = 45$
Total number of times the dice rolled are $400 t i m e s$
$\Rightarrow n (S) = 400$
We know that $P (E) = \frac{n (E)}{n (S)}$
Substituting the values we get, $P (E) = \frac{45}{400} = 0.1125$
Step 2: Find the probability of getting a sum of more than $10$ on the top:
Let $E_{1}$ be the event of getting a sum of more than $10$ on the top of two dice.
From the frequency distribution table, $n (E_{1}) = 18 + 5 = 23$
Total number of times the dice rolled are $400 t i m e s$
$\Rightarrow n (S) = 400$
We know that $P (E) = \frac{n (E)}{n (S)}$
Substituting the values we get, $P (E) = \frac{23}{400}$
Step 3: Find the probability of getting a sum of between $5 a n d 10$ on the top:
Let $E_{2}$ be the event of getting a sum of between $5 a n d 10$ on the top of two dice.
From the frequency distribution table, $n (E_{2}) = 62 + 65 + 60 + 43 = 230$
Total number of times the dice rolled are $400 t i m e s$
$\Rightarrow n (S) = 400$
We know that $P (E) = \frac{n (E)}{n (S)}$
Substituting the values we get, $P (E) = \frac{230}{400} = \frac{23}{40}$
Hence, the probability of getting a sum of $5$ , more than $10$ and between $5 a n d 10$ are $\frac{9}{80}, \frac{23}{400} a n d \frac{23}{40}$ respectively.

Suggest Corrections

Similar questions

Q. (i) Complete the following table:

Event: 'Sum on $2$ dice'	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	$10$	$11$	$12$
Probability	$\frac{1}{36}$						$\frac{5}{36}$				$\frac{1}{36}$

(ii) A student argues that there are

11

possible outcomes

2, 3, 4, 5, 6, 7, 8, 9, 10, 11

and

12

. Therefore, each of them has a probability

\frac{1}{11}

. Do you agree with this argument?
Justify your answer.

Prove that:
(i) $\frac{1}{3 + \sqrt{7}} + \frac{1}{\sqrt{7} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{3}} + \frac{1}{\sqrt{3} + 1} = 1$
(ii) $\frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \frac{1}{\sqrt{4} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{6}} + \frac{1}{\sqrt{6} + \sqrt{7}} + \frac{1}{\sqrt{7} + \sqrt{8}} + \frac{1}{\sqrt{8} + \sqrt{9}} = 2$

Q. Solve the following equations.
(1) 17p − 2 = 49
(2) 2m + 7 = 9
(3) 3x + 12 = 2x − 4
(4) 5(x − 3) = 3(x + 2)

(5)

\frac{9 x}{8}

+ 1 = 10

(6)

\frac{y}{7}

\frac{y - 4}{3}

= 2

(7) 13x − 5 =

\frac{3}{2}

(8) 3(y + 8) = 10(y − 4) + 8

(9)

\frac{x - 9}{x - 5}

\frac{5}{7}

(10)

\frac{y - 4}{3}

+ 3y = 4

(11)

\frac{b + (b + 1) + (b + 2)}{4}

= 21

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 $12$ ?

Q. Simplify:
(i)