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maw [93]
3 years ago
11

You and a friend are rolling two 6-sided number cubes and placing bets on the sum of the outcome. is it better to place a bet on

a sum of 3, 4, 7, or 11?
Mathematics
2 answers:
mote1985 [20]3 years ago
6 0

Answer:

It is better to bet on 7.

Step-by-step explanation:

It is given that,

We are rolling two 6-sided numbered cubes and placing bets on the 'sum of the outcome'.

So, the pair of numbers will be the possible options for the outcome to be their sum.

Now, the possibility to have the sum of 3, 4, 7 and 11 are:

3 = 1+2 i.e. only one pair is the possibility

4 = 1+3 = 2+2 i.e there are two pairs as the possibility

7 =  1+6 = 2+5 = 3+4 i.e. there are three possible pairs

11 = 5+6 i.e. only one pair is the possibility

So, the maximum possibility of a sum to be the outcome is when the sum = 7 i.e. there are 3 possible pairs for the sum to be 7.

Hence, 7 has the highest possibility to become the outcome.

So, It is better to bet on 7.

Zina [86]3 years ago
4 0
To solve for this, we need to figure out the sum of any two numbers on the dice that will become a pair factor for the listed answers.
What I mean by that is that we need to find the probability between how many two numbers will sum up to the answer.

3 can be 1 + 2, that's it. 3 has a probability of (1).
4 has a sum of 1 + 3, and 2 + 2, that's it. 4 has a probability of (2).
7 has a sum of 1 + 6, 2 + 5, and 3 + 4. 7 has a probability of (3).
11 has a sum of 5 + 6, that's it. 11 has a probability of (1).

The numbers in parenthesis are the probabilities.
7 has the highest number, so 7 has the highest probability.

Your answer is: 7.

I hope this helps!
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3 years ago
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denis-greek [22]

<u>Answer:</u>

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<u>Step-by-step explanation:</u>

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Second, get the 12 on the right side of the equal sign by adding a -12 to each side

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<u>AND</u>

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8 0
4 years ago
<img src="https://tex.z-dn.net/?f=%5Csqrt288x4y9" id="TexFormula1" title="\sqrt288x4y9" alt="\sqrt288x4y9" align="absmiddle" cla
sasho [114]

Here's what I got, hope it helps.

5 0
3 years ago
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