Question

A die is tossed 120 times and the outcomes are recorded as follows:
Outcomes
1
Even no.<6
Odd no.>1
6

Frequency
25
40
35
20

Find the probability of getting :
an even number.

an odd number greater than 1.

getting 1.​

Answer 1

Answer:

I hope you got your answer

Answer 2

Given:

A die is tossed 120 times and the outcomes are recorded as follows:

Outcomes        Frequency

1                              25

Even no.<6            40  

Odd no.>1              35

6                             20

To find: The probability of getting :

(a) an even number.

(b) an odd number greater than 1.

(c) getting 1.​

Solution:

The total number of possible outcomes for each case is 120.

(a)

The number of favourable outcomes for an even number must be the sum of 40 and 20. Hence, the probability can be written as follows.

$P(E) = \frac{40+20}{120}$

         $= \frac{60}{120}$

         $= \frac{1}{2}$

(b)

The number of favourable outcomes for an odd number greater than 1 must be 35. Hence, the probability can be written as follows.

$P(E) = \frac{35}{120}$

         $= \frac{7}{24}$

(c)

The number of favourable outcomes for getting 1 must be 25. Hence, the probability can be written as follows.

$P(E) = \frac{25}{120}$

         $=\frac{5}{24}$

Therefore, the probability of getting :

(a) an even number is 1/2.

(b) an odd number greater than 1 is 7/24.

(c) getting 1 is 5/24​.