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Q: How many leaves on a tree diagram are needed to represent all possible combinations of tossing a coin and rolling a die?

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48

The answer is 12 APEX ✨

There are 8 possible combinations that would produce a 5 or 9 and 8/36=2/9.

The probability of rolling a 9 with two dice is 4/36 or 1/9. There are 36 possible combinations and a 9 is obtained by rolls of 3,6; 4,5; 6,3; & 5,4.

Of the 36 possible combinations rolling two dice there are 2 combinations that add up to 11 so the odds are 18:1

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48

The answer is 12 APEX ✨

There are 8 possible combinations that would produce a 5 or 9 and 8/36=2/9.

The probability of rolling a 9 with two dice is 4/36 or 1/9. There are 36 possible combinations and a 9 is obtained by rolls of 3,6; 4,5; 6,3; & 5,4.

Of the 36 possible combinations rolling two dice there are 2 combinations that add up to 11 so the odds are 18:1

To find the probability that when rolling a die and tossing a coin, your will obtain an odd on the die OR a heads on the coin, use the addition rule, which is: P(A) + P(B) - P(A and B) = P(A or B In this example, event A is tossing heads on the coin, and event B is rolling odd on the die. What you are trying to solve is actually A U B (A union B) First the sample set of all 12 possible combinations: S={H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6} The 6 heads possible combinations are: A={H1, H2, H3, H4, H5, H6| The 6 odd number possible combinations are: B={H1, H3, H5, T1, T3, T5} The 3 combinations these sets have in common, A intersect B: A n B= {H1. H3, H5} There are 12 possible combinations and 6 of those include a heads on the coin. This is 6/12. There are 12 possible combinations and 6 of those include an odd on the die. This is 6/12. There are 12 possible combinations and 3 of those include both an odd on the die, and a heads on the coin. This is 3/12. 6/12 + 6/12 - 3/12 = 9/12 Simplify the above number to 3/4, which is the correct answer to this question. If you draw a Venn diagram, you will see that the set consisting of tails and evens {T2, T4, T6} falls outside the circles. The diagram makes it easy to see that 9 of the 12 possible combinations fall inside the circle, and 3 of the 12 fall outside. Hope this helps someone. I solidified the information for myself by writing it!

The chance is 1/36. (There are 36 possible combinations for two 6-sided dice, but only 18 separate combinations when the dice are not considered seperately.)

11 = 6+5 is the only solution, so there are two combinations first dice : 6, second dice : 5 first dice : 5, second dice : 6

If you have 12 possible numbers with multiple combinations then you should start out with making all the possible combinations; you will find theyre 20. Theyre four numbers out of the twleve that can be divisible by three; 3, 6, 9, and 12. There are 7 combinations where the combinations can equal those four numbers. So the odds of getting a sum divisible by three is 7/20.

Te probability of rolling a sum of 7 with two fair dice is 6 in 36, or 1 in 6, or about 0.1667.Of all the possible combinations of two dice, the sum of 7 has the highest probability, with the other combinations decreasing down to 2 and 12, with probabilities of 1 in 36, or about 0.0278.

The combinations that produce a 10 are (4,6),(5,5) and (6,4), and there are 36 combinations (6 x 6) possible, so 3/36 = 1/12. So one chance in 12. The probability is 0.0833.

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