Toss 3 coin and 2 die find the probability of appearing 3 heads and sum of a dot = 12

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To find the probability of obtaining 3 heads when tossing 3 coins and a sum of the dots equal to 12 when rolling 2 dice, we need to calculate the probability of each event and then multiply them together since the events are independent.
The probability of getting 3 heads when tossing 3 coins is given by the binomial probability formula. Since each coin has 2 possible outcomes (heads or tails) and we want to obtain 3 heads, the probability can be calculated as:

P (3 heads) = (\binom{3}{3}) \cdot {(\frac{1}{2})}^{3} = \frac{1}{8}

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The probability of obtaining a sum of the dots equal to 12 when rolling 2 dice can be found by determining the favorable outcomes and dividing it by the total possible outcomes. To count the favorable outcomes, we need to determine the combinations of dice rolls that result in a sum of 12:
- (6, 6)
There is only one favorable outcome. The total number of possible outcomes when rolling 2 dice is 6 * 6 = 36.
Therefore, the probability of obtaining a sum of 12 when rolling 2 dice is:

P (sum = 12) = \frac{1}{36}

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Since the events of getting 3 heads and a sum of 12 are independent, we can multiply their probabilities:

P (3 heads and sum = 12) = P (3 heads) \cdot P (sum = 12) = \frac{1}{8} \cdot \frac{1}{36} = \frac{1}{288}

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Thus, the probability of obtaining 3 heads when tossing 3 coins and a sum of the dots equal to 12 when rolling 2 dice is

\frac{1}{288}

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Toss 3 coin and 2 die find the

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