Throw a coin and a dice.

Probability Level 2

I have a dice and a coin. And I will throw them.

Let $a$ to the probability of getting a front face of a coin, and $b$ to the probability of getting a product $8$ , or a sum $5$ . ( I will throw a die twice, and a coin once. And the dice is a regular hexahedron. And the dice has 1 ~ 6 eyes. And, the answer is 21 decimal digits.)

Find $\displaystyle \frac { a } { b } + \frac { b } { a }$ .

The answer is 3.333333333333333333333.

1 solution

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Feb 26, 2021

A coin has one front face and one back face, so the probability $a$ is $\displaystyle \frac { 1 } { 2 }$ .

And a dice has 6 faces because the problem has said the dice is regular hexahedron.

So, $b$ equals to $\displaystyle \cancel { ( 1, 8 ) }, ( 2, 4 ), ( 4, 2 ), \cancel { ( 8, 1 ) } \rightarrow \frac { 2 } { 36 }, ( 1, 4 ), ( 2, 3 ), ( 3, 2 ), ( 4, 1 ) \rightarrow \frac { 4 } { 36 } \Rightarrow \frac { 2 } { 36 } + \frac { 4 } { 36 } = \frac { 6 } { 36 } = \frac { 1 } { 6 }$ .

Then, $\displaystyle \displaystyle \frac { a } { b } + \frac { b } { a } = \frac { \frac { 1 } { 2 } } { \frac { 1 } { 6 } } + \frac { \frac { 1 } { 6 } } { \frac { 1 } { 2 } } = \frac { 1 } { 2 } \div \frac { 1 } { 6 } + \frac { 1 } { 6 } \div \frac { 1 } { 2 } = \frac { 1 } { \cancel { 2 } _ { 1 } } \times \cancel { 6 } _ { 3 } + \frac { 1 } { \cancel { 6 } _ { 3 } } \times \cancel { 2 } _ { 1 } = 3 + \frac { 1 } { 3 } = 3.333333333\cdots$ .

Throw a coin and a dice.

The answer is 3.333333333333333333333.

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