Question

Probability of getting atleast one odd number when a die is thrown thrice

Answer 1

When you roll a dice you have 1/2 of the probability to get an odd number and 1/2 to get an even one.

Probability= (number of cases that you get an odd at least once) / ( all the possible cases)

The probability to drawn 3 even numbs is: 1/2*1/2*1/2=1/8.

Therefore the probability to get at least one number odd is 1–1/8 = 7/8. ( 1 — the probability to get 3 evens, which is 1/8 ).

Answer 2

Hi there!

Given :- A die is tossed thrice!

P (Odd no. once) = P (Once) + P (Twice) + P (Thrice)

= 1 - Probability of an odd no. 0 times.

= 1 - Probability of an odd no. all 3 times.

Let the event be :-

→ O : Getting an Odd number
→ E : Getting an Even number

Now,
Required probability = 1 - P(EEE)

So, P(EEE) = P(E) × P(E | E) × P(EE | E)

♦ P (Getting an Even number) = P(E) = 3 / 6 = 1 / 2

♦ P (Getting an Even no. 2nd time) = P(E | E) = 3 / 6 = 1 / 2

♦ P (Getting an Even no. 3rd time) = P(EE | E) = 3 / 6 = 1 / 2

Thus,

P(EEE) = P(E) × P(E | E) × P(EE | E)

= 1 / 2 × 1 / 2 × 1 / 2

= 1 / 8

∴ P (Getting an Odd no. at-least once) = 1 - P(EEE)

= 1 - 1 / 8

= 8 - 1 / 8

= 7 / 8

Hence, The required probability is :- 7 / 8

Hope it helps! :)