Question

In throwing 3 dice the probability that at least 2 of the three numbers obtained are same

Answer 1

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Here's the answer:

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• Find the probability that all show different numbers--- P(E')

• Then find the probability that at least two dice have to show the same number --- P(E)

°•° P(E) + P(E') = 1
(°•° Sum of probabilities = 1 )

=> Required probability P(E) = 1 - P(E')

• No. of total outcomes when n dice are rolled = 6^n

• Probability of occurrence an Event
= No. of Favourable Outcomes of that Event / Total No. of outcomes

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SOLUTION :

Each dice has 6 faces & getting any number will have equal probability.

Here we have 3 unbiased dice

Let S be the sample space
n(S) - No. of total outcomes when 3 dice are rolled
n(S) = 6³ = 6×6×6 = 216

STEP-1 : Find P(E')

E' be the Event that no dice have the same number

••• For 3 dice ,

° The first dice can take any of 6 number when theft first dice is rolled.

° Now, there are only 5 options left for second dice and only 4 left for the third dice so as to satisfy the Event E'.

n(E') - total number of viable options to satisfy Event E

n(E') = 6 × 5 × 4 = 120

Probability that no dice have the same number P(E') = n(E') / n(S)

P(E') = 120/216 = 5/9

•°• The probability that all 3 dice show a different number is 5/9

STEP -2 : Find P(E)

E be the Event that at least two dice have to show same number

P(E) = 1 - P(E') = 1 - 5/9 = 4/9

The probability that at least 2 of them show the same number is 4/9

•°• Required probability = 4/9

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Hope it helps