Question

1 In how many ways can a family of 9 divide itself into 3 groups so that each group contains 3 persons?

2 Four digits numbers are to be formed using any of the digits 1, 2, 3, 4, 5, 6. If no repetition of digit is allowed how many 4-digit even numbers can be formed?

3 A club consist of 10 men and 5 women, in how many ways can a committee of 6 consisting of 4 men and 2 women be chosen?

4 Let A and B be any two events defined on the same sample space. Suppose P(A) = 0.3 and P(A B) = 0.6. Find P(B) such that A and B are independent.

5 Let A and B be any two events defined on the same sample spaceSuppose P(A) = 0.3 and P(A B) = 0.6. Find P(B) such that A and B are mutually exclusive.

Answer 1

Answer on Question #59918 – Math – Statistics and Probability

Question

1. In how many ways can a family of 9 divide itself into 3 groups so that each group contains 3 persons?

Solution

We are seeking for unordered partitions $r_1 = 3, r_2 = 3, r_3 = 3$.

The number of unordered partitions is

$$\frac{9!}{3! \cdot 3! \cdot 3!} \cdot \frac{1}{3!} = \frac{9!}{6^4} = 280$$

(since the three parts contain the same number of objects).

The same relationship that exists between permutation and combination exist between ordered and unordered partition of a set.

**Answer:** 280.

Question

2. Four digits numbers are to be formed using any of the digits 1, 2, 3, 4, 5, 6. If no repetition of digit is allowed how many 4-digit even numbers can be formed?

Solution

This is permutation of 4 digits from 6 digits. Therefore, the number of different permutations is

$$^6 P_4 = \frac{6!}{(6 - 4)!} = \frac{6!}{(2)!} = 6 \cdot 5 \cdot 4 \cdot 3 = 360.$$

Alternatively, we can reason as follows:

The first digit to be selected can be any of the six given digits, so $n_1 = 6$. The second digit to be selected can be any of the remaining 5 digits (since no reparation is allowed) so $n_2 = 5$. Similarly, $n_3 = 4$ and $n_4 = 3$. Thus, the answer is

$$n_1 \cdot n_2 \cdot n_3 \cdot n_4 = 6 \cdot 5 \cdot 4 \cdot 3 = 360.$$

**Answer:** 360.

Question

3. A club consists of 10 men and 5 women, in how many ways can a committee of 6 consisting of 4 men and 2 women be chosen?

Solution

The 4 men can be chosen from the 10 men in $^{10}C_4$ ways, the 2 women can be chosen from the 5 women in $^{5}C_2$ ways. Hence the committee can be chosen in (by the fundamental principle of counting)

$$^{10}C_4 \cdot ^5C_2 = \frac{10!}{6! \cdot 4!} \cdot \frac{5!}{2! \cdot 3!} = 2100 \text{ ways}$$

Answer: 2100 ways.

Question

4 Let A and B be any two events defined on the same sample space. Suppose $P(A) = 0.3$ and $P(A \ B) = 0.6$. Find $P(B)$ such that A and B are independent.

Solution

For independent events we obtain

$$P(A \cap B) = P(A)P(B)$$

If $P(A \cap B) = 0.6$, then $P(B) = \frac{P(A \cap B)}{P(A)} = \frac{0.6}{0.3} = 2$, which is impossible, because $P(B) \leq 1$ has to be true.

If $P(A \cup B) = 0.6$, then by the rule of addition,

$$P(A \cup B) = P(A) + P(B) - P(A \cap B) = P(A) + P(B) - P(A)P(B).$$ $$P(B) = \frac{P(A \cup B) - P(A)}{1 - P(A)} = \frac{0.6 - 0.3}{1 - 0.3} = \frac{3}{7}.$$

Answer: $\frac{3}{7}$.

Question

5 Let A and B be any two events defined on the same sample space. Suppose $P(A) = 0.3$ and $P(A \ B) = 0.6$. Find $P(B)$ such that A and B are mutually exclusive.

Solution

By the rule of addition,

$$P(A \cup B) = P(A) + P(B) - P(A \cap B).$$

If A and B are mutually exclusive, then $P(A \cap B) = 0$. It follows that

$$P(B) = P(A \cup B) - P(A) = 0.6 - 0.3 = 0.3.$$

Answer: 0.3.

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