Get Answers to all your Questions

header-bg qa

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

Answers (1)

\text{Probability }=\frac{\text {Favourable outcomes}}{\text {Total number of events}}

Here, total events  = 14 + 30 + 42 + 55 + 72 + 75 + 70 + 53 + 46 + 28 +15 = 500

(i) probability of getting a sum = 3

Favourable events = 30

\text{P[of getting sum 3]}=\frac{30}{500}=0.06

(ii) probability of getting a sum more than 10

\text{Favourable events}= 28 + 15 = 43

\text{P[of getting sum 10]}=\frac{43}{500}=0.086

(iii) probability of getting a sum less than or equal to 5

Favourable events = 14 + 30 + 42 + 55

\text{p[sum less than or equal to 5]}=\frac{141}{500}=\frac{28.2}{100}=2.282

(iv) probability of getting a sum between 8 and 12

Favourable events = 53 + 46 + 28 = 127

\text{p[sum between 8 and 12]}=\frac{127}{500}=0.254

 

Posted by

infoexpert23

View full answer