Question

A die is thrown once. Find the probability of getting (i) a prime number; (ii) a number lying between 2 and 6; (iii) an odd number.

Answer 1

SOLUTION
Possible numbers of events on throwing a dice = 6
Numbers on dice = 1,2,3,4,5 and 6 

(i) Prime numbers = 2, 3 and 5
Favourable number of events = 3
Probability that it will be a prime number = 3/6 = 1/2

(ii) Numbers lying between 2 and 6 = 3, 4 and 5
Favourable number of events = 3
Probability that a number between 2 and 6 = 3/6 = 1/2

(iii) Odd numbers = 1, 3 and 5
Favourable number of events = 3
Probability that it will be an odd number = 3/6 = 1/2

Answer 2

Answer:

The probabilities are (i)1/2 (ii)1/2 (iii)1/2

Step-by-step explanation:

As per the data given in the question,

we have,

No. of Possible outcomes on throwing a dice(1,2,3,4,5,6) = 6

so,

(i) a prime number

No. of prime no.s(2,3,5)=3

∴Probability = $\frac{favourable \:outcomes}{total \: outcomes}$ = $\frac{3}{6} =\frac{1}{2}$

(i) a number lying between 2 and 6

No. of numbers lying between 2 and 6(3,4,5)=3

∴Probability = $\frac{favourable \:outcomes}{total \: outcomes}$ = $\frac{3}{6} =\frac{1}{2}$

(i) an odd number

No. of odd no.s(1,3,5)=3

∴Probability = $\frac{favourable \:outcomes}{total \: outcomes}$ = $\frac{3}{6} =\frac{1}{2}$

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