I misunderstood the question. I thought, it needs either 1 in all the trials or 2 on all the trials not in every trial we have two choices with 1 or 2. Thanks

1st time: Probability of getting 1 OR 2 when die is tossed: 1/6 + 1/6 = 2/6 = 1/3
2nd time: Probability of getting 1 OR 2 when die is tossed: 1/6 + 1/6 = 1/3
3rd time: Probability of getting 1 OR 2 when die is tossed: 1/6 + 1/6 = 1/3

so 1st time AND 2nd time AND 3rd time: 1/3 * 1/3 * 1/3 = 1/27

Look:

in every toss, there are 6 probable outcomes, so in 3 tosses total outcome = 6*6*6 =216

As, Probability = (# of possible outcome) / (# of total outcome =216)

For possible outcome: imagine this as we have 3 boxes each of which can be filled with either 1 or 2; __ __ __

1st box has two candidates: 1 and 2; so as the 2nd and 3rd ==

2 2 2 = 8 = #of possible outcomes

Ans: P = \(\frac{8}{216}\) = \(\frac{1 }{ 27}\)
You got it !!!!

Given that A fair six-sided die with faces numbered 1 through 6 is tossed 3 times and We need to find what is the probability of getting a 1 or a 2 on all three tosses?

As we are rolling a dice thrice => Number of cases = \(6^3\) = 216

We need to get a 1 or a 2 in all three rolls.

Probability of getting 1 or 2 in one roll = \(\frac{2}{6}\) (as there are two favorable outcomes 1 or 2 out of 6) = \(\frac{1}{3}\)

=> Probability of getting a 1 or a 2 on all three tosses = \(\frac{1}{3}\) * \(\frac{1}{3}\) * \(\frac{1}{3}\) = \(\frac{1}{27}\)

So, Answer will be \(\frac{1}{27}\)
Hope it helps!

Watch the following video to learn How to Solve Dice Rolling Probability Problems

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