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A dice is thrown 3 times and the sum of the 3 numbers thrown is 15. Then, what is the probability that the first throw was a four?

My Attempt:

The sample space for the above problem, $S = 6^3$.

Combinations of thrown numbers such that their sum is $15$ and the first digit is $4 = [\{4, 5, 6\}, \{4, 6, 5\}]$

Thus, the total number of possible sets are 2. Thus, the probability for the required event (say $A$), $$P(A) = \frac{2}{216}$$ $$\therefore P(A) = \frac{1}{108}$$

In my book, the answer given is $\frac{1}{5}$. How's is my answer?

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2 Answers 2

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Since the sum of the 3 numbers thrown is 15 it follows that the sample space is given by 10 elements: 15=6+6+3 ($3!/2!=3$ ways), 15=6+5+4 (3!=6 ways), 15=5+5+5 ($3!/3!=1$ way).

There are two cases when $4$ is the first value 4,5,6 and 4,6,5.

Therefore the probability is $p=2/10=1/5$.

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You have to consider all the possible ways in which you can obtain $15$.

Those possibilities are:

$\{3, 6, 6 \}$, $\{6, 3, 6 \}$, $\{6, 6, 3 \}$

$\{5, 5, 5 \}$

$\{4, 5, 6 \}$, $\{4, 6, 5 \}$, $\{5, 4, 6 \}$, $\{5, 6, 4 \}$, $\{6, 4, 5 \}$, $\{, 5, 4 \}$

Total = $10$

There are only $2$ cases in which $4$ is the first element, so

$$P = \frac{2}{10} = \frac{1}{5}$$

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