Q: How many leaves on a tree diagram are needed to represent all possible combinations tossing a coin 5 time?

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The answer is 12 APEX ✨

There are 36 possible combinations. Eleven of them have at least one four in it. That means it is 11 over 36, which is a 30.55% chance.

It is used for lots of things such as finding out the total possible outcomes of tossing coins. You find the line that corresponds with how many coins you toss and add all the numbers in that line to get the number of possible outcomes also you can use it to find combinations and permutations and triangular numbers

You cannot. The tree diagram for tossing 4 coins has 16 branches. So if that is done 96 times, you will have a tree with 1696 branches which is approx 4 trillion googol branches.

Is possible.

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Answer is 16 on apex. Trust me

The answer is 12 APEX ✨

It is used to represent one of the two possible outcomes of tossing a coin.

18 different combinations. When a coin is tossed twice there are four possible outcomes, (H,H), (H,T), (T,H) and (T,T) considering the order in which they appear (first or second). But if we are talking of combinations of the two individual events, then the order in which they come out is not considered. So for this case the number of combinations is three: (H,H), (H,T) and (T,T). For the case of tossing a die once there are six possible events. The number of different combinations when tossing a coin twice and a die once is: 3x6 = 18 different combinations.

There are 25 or 32 possible outcomes can you get by tossing 5 coins.

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!

There are 36 possible combinations. Eleven of them have at least one four in it. That means it is 11 over 36, which is a 30.55% chance.

There are 23 = 8 possible outcomes.

It is used for lots of things such as finding out the total possible outcomes of tossing coins. You find the line that corresponds with how many coins you toss and add all the numbers in that line to get the number of possible outcomes also you can use it to find combinations and permutations and triangular numbers

You cannot. The tree diagram for tossing 4 coins has 16 branches. So if that is done 96 times, you will have a tree with 1696 branches which is approx 4 trillion googol branches.

There are 26 = 64 possible outcomes.

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