Question

A dice is thrown 5 Times . What is the probability of getting sum 25 ?

Answer 1

Answer on Question #64180 – Math – Statistics and Probability

Question

A dice is thrown 5 times. What is the probability of getting sum 25?

Solution

Method 1

Let $A$ be the event of getting the sum of 25, $N$ is the number of all possibilities and $K$ is the number of possibilities satisfying $A$.

Let's use the classical definition of probability:

$$P(A) = \frac{K}{N}.$$

$N$ is equal to the number of all possible variants to choose the outcome five times:

$$N = 6^5 = 7776.$$

We can receive $K$ as the number of nonnegative integer solutions to the equation

$$k_1 + k_2 + k_3 + k_4 + k_5 = 25, \text{ where } 1 \leq k_i \leq 6, i = 1, 2, 3, 4, 5.$$

Let $k_1 = m_1 + 1$, $k_2 = m_2 + 1$, $k_3 = m_3 + 1$, $k_4 = m_4 + 1$, $k_5 = m_5 + 1$, where $0 \leq m_i \leq 5, i = 1, 2, 3, 4, 5$.

Then

$$k_1 + k_2 + k_3 + k_4 + k_5 = m_1 + 1 + m_2 + 1 + m_3 + 1 + m_4 + 1 + m_5 + 1 = 25,$$

hence

$$m_1 + m_2 + m_3 + m_4 + m_5 = 20, \quad 0 \leq m_i \leq 5, \quad i = 1, 2, 3, 4, 5.$$

Let $P_1 \Leftrightarrow m_1 > 5 \Leftrightarrow m_1 \geq 6$, $P_2 \Leftrightarrow m_2 > 5 \Leftrightarrow m_2 \geq 6$, $P_3 \Leftrightarrow m_3 > 5 \Leftrightarrow m_3 \geq 6$, $P_4 \Leftrightarrow m_4 > 5 \Leftrightarrow m_4 \geq 6$, $P_5 \Leftrightarrow m_5 > 5 \Leftrightarrow m_5 \geq 6$.

There are $|S| = \binom{20 + 5 - 1}{5 - 1} = \binom{24}{4} = \frac{24 \cdot 23 \cdot 22 \cdot 21}{4 \cdot 3 \cdot 2} = 23 \cdot 22 \cdot 21 = 10626$ non-negative solutions to the equation $m_1 + m_2 + m_3 + m_4 + m_5 = 20, m_i \geq 0$.

Solutions in $P_1$ correspond to non-negative integer solutions of

$l_{1} + m_{2} + m_{3} + m_{4} + m_{5} = 20 - 6$ after substitution $m_{1} = l_{1} + 6$, $l_{1} \geq 0$, the answer is $\binom{20-6+5-1}{5-1} = \binom{18}{4} = \frac{18 \cdot 17 \cdot 16 \cdot 15}{4 \cdot 3 \cdot 2} = 3 \cdot 17 \cdot 4 \cdot 15 = 3060$.

Solutions in $P_{1} \cap P_{2}$ correspond to non-negative integer solutions of

$l_{1} + l_{2} + m_{3} + m_{4} + m_{5} = 20 - 6 - 6$ after substitutions

$m_{1} = l_{1} + 6, m_{2} = l_{2} + 6, l_{1} \geq 0, l_{2} \geq 0$, the answer is

$$\binom{20 - 6 - 6 + 5 - 1}{5 - 1} = \binom{12}{4} = \frac{12 \cdot 11 \cdot 10 \cdot 9}{4 \cdot 3 \cdot 2} = 11 \cdot 5 \cdot 9 = 495.$$

Solutions in $P_{1} \cap P_{2} \cap P_{3}$ correspond to non-negative integer solutions of

$l_{1} + l_{2} + l_{3} + m_{4} + m_{5} = 20 - 6 - 6 - 6$ after substitutions

$m_{1} = l_{1} + 6, m_{2} = l_{2} + 6, m_{3} = l_{3} + 6, l_{1} \geq 0, l_{2} \geq 0, l_{3} \geq 0,$

the answer is $\binom{20 - 6 - 6 - 6 + 5 - 1}{5 - 1} = \binom{6}{4} = \frac{6 \cdot 5 \cdot 4!}{4! \cdot 2!} = 15.$

Applying inclusion-exclusion formula, we have

$$\begin{array}{l} \left| P_{1} \cup P_{2} \cup P_{3} \cup P_{4} \cup P_{5} \right| = \sum_{i=1}^{5} \left| P_{i} \right| - \sum_{1 \leq i < j \leq 5} \left| P_{i} \cap P_{j} \right| + \sum_{1 \leq i < j < k \leq 5} \left| P_{i} \cap P_{j} \cap P_{k} \right| = \\ = 5 \cdot 3060 - 10 \cdot 495 + 10 \cdot 15 = 10500. \end{array}$$

We are interested in

$$\begin{array}{l} K = \left| \bar{P}_{1} \cap \bar{P}_{2} \cap \bar{P}_{3} \cap \bar{P}_{4} \cap \bar{P}_{5} \right| = |S| - \left| P_{1} \cup P_{2} \cup P_{3} \cup P_{4} \cup P_{5} \right| = 10626 - \\ -10500 = 126. \end{array}$$

Finally

$$P(A) = \frac{K}{N} = \frac{126}{7776}.$$

Method 2

We can receive $K$ by writing down and counting all suitable situations sorted by minimal number received:

1) $k_{1} = 1, k_{2} = 6, k_{3} = 6, k_{4} = 6, k_{5} = 6$. There are 5 ways depending on a number thrown.

2) $k_{1} = 2, k_{2} = 5, k_{3} = 6, k_{4} = 6, k_{5} = 6$. There are 5 ways to locate "2" and 4 possibilities to get "5" at once. So we have $5 \cdot 4 = 20$ variants.

3)

A) $k_{1} = 3, k_{2} = 4, k_{3} = 6, k_{4} = 6, k_{5} = 6.$ (Similarly, 20 variants)

B) $k_{1} = 3, k_{2} = 5, k_{3} = 5, k_{4} = 6, k_{5} = 6$. In this case there are 5 possibilities to locate "3", 4 possibilities to get "5" for the first time and 3 possibilities to get "5" for the second time.

We get $5 \cdot 4 \cdot 3 = 60$ variants.

4) $k_{1} = 4, k_{2} = 5, k_{3} = 5, k_{4} = 5, k_{5} = 6.$ (20 variants)

5) $k_{1} = 5, k_{2} = 5, k_{3} = 5, k_{4} = 5, k_{5} = 5$ (1 variant)

Variant of getting "6" five times is impossible, because the sum of numbers will be 30 then.

Now

$$K = 5 + 20 + 20 + 60 + 20 + 1 = 126,$$

Finally

$$P(A) = \frac{K}{N} = \frac{126}{7776}.$$

Answer: $\frac{126}{7776}$.

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