A die is tossed 3 times, what is the probability of 1 five

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A die is having six sides with six different numbering

Even number in the die are {2,4,6}

The odd number in the die are {1,3,5}

 \(Probability~of~an~Event =\frac{Number~of~desired~Events }{Total~number~of~Events}\)

The probability of occurring an odd number is = 3/6 = 1/2

The probability of occurring an even number is = 3/6 = 1/2

When die is tossed three times and getting an odd number at least once is given by

p(at least one odd number) = 1 - p(no odd number in all three tosses)

\(p(at~least~one~odd~number) =1- \frac{1}{8}\)

p(no odd number in all three tosses) = p(even) × p(even) × p(even)

\(p(no~odd~number~in~all~three~tosses) = \frac{1}{2}×\frac{1}{2}×\frac{1}{2}\)

\(p(no~odd~number~in~all~three~tosses) = \frac{1}{8}\)

\(p(at~least~one~odd~number) =\frac{7}{8}\)

Hence option (2) is correct 

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Since you're performing multiple experiments each with exactly two outcomes (the dice is a five vs. the dice is not a five), this is a classic example of a Bernoulli trial.

The standard formula to calculate the probability is then to use (from the Binomial distribution):

$$ P\{\text{that A occurs exactly k times in n trials}\} = \left( {\begin{array}{*{20}c} n \\ k \\ \end{array}} \right) p^k (1 - p)^{n-k} $$

where event $A$ is the trial and $p$ is the probability for the success of a single trial ($\frac{1}{6}$).

In your case, you want to know the probability of it happening at least twice, so you need to sum the chances for it happening exactly twice and exactly three times:

$$ P\{\text{that A occurs exactly 2 times in 3 trials}\} + P\{\text{that A occurs exactly 3 times in 3 trials}\} $$

Therefore, $$ \begin{align} P&= \left( {\begin{array}{*{20}c} 3 \\ 2 \\ \end{array}} \right) p^2 (1 - p)^{(3-2)} + \left( {\begin{array}{*{20}c} 3 \\ 3 \\ \end{array}} \right) p^3 (1 - p)^{(3-3)} \\ &= \left( {\begin{array}{*{20}c} 3 \\ 2 \\ \end{array}} \right) \cdot \frac{1}{6}^2 \cdot \frac{5}{6}^{1} + \left( {\begin{array}{*{20}c} 3 \\ 3 \\ \end{array}} \right) \cdot \frac{1}{6}^3 \cdot \frac{5}{6}^{0}\\ &= 3 \cdot \frac{1}{36} \cdot \frac{5}{6} + 1 \cdot \frac{1}{216} \cdot 1\\ &= \frac{15}{216} + \frac{1}{216}\\ &= \frac{16}{216}\\ &= \frac{2}{27} = 0,074... = 7,41 \% \\ \end{align} $$

As you can see, this method of calculation can unfortunately become very tedious for larger examples, but it has the advantage of being a consistent approach for any problems surrounding Bernoulli trials. The intuitive approaches mentioned in some of the other answers can be quicker for small cases where all the possible outcomes can be manually considered.