In a group there are 3 women and 3 men . 4 people are selected at random from this group . Find the probability that 3 women and 1 man or 1 women and 3 men are selected .

Answer

To solve the problem step by step, we will calculate the probability of selecting either 3 women and 1 man or 1 woman and 3 men from a group of 3 women and 3 men.

### Step 1: Determine the total number of ways to select 4 people from the group.
The total number of people in the group is 6 (3 women + 3 men). We need to select 4 people from these 6.

\[ \text{Total ways} = \binom{6}{4} \]

### Step 2: Calculate \(\binom{6}{4}\).
Using the formula for combinations, \(\binom{n}{r} = \frac{n!}{r!(n-r)!}\):

\[ \binom{6}{4} = \frac{6!}{4! \cdot (6-4)!} = \frac{6!}{4! \cdot 2!} = \frac{6 \times 5}{2 \times 1} = 15 \]

### Step 3: Calculate the probability of selecting 3 women and 1 man.
Let event A be selecting 3 women and 1 man. The number of ways to select 3 women from 3 is:

\[ \binom{3}{3} = 1 \]

The number of ways to select 1 man from 3 is:

\[ \binom{3}{1} = 3 \]

Thus, the total number of favorable outcomes for event A is:

\[ \text{Favorable outcomes for A} = \binom{3}{3} \cdot \binom{3}{1} = 1 \cdot 3 = 3 \]

### Step 4: Calculate the probability of selecting 1 woman and 3 men.
Let event B be selecting 1 woman and 3 men. The number of ways to select 1 woman from 3 is:

\[ \binom{3}{1} = 3 \]

The number of ways to select 3 men from 3 is:

\[ \binom{3}{3} = 1 \]

Thus, the total number of favorable outcomes for event B is:

\[ \text{Favorable outcomes for B} = \binom{3}{1} \cdot \binom{3}{3} = 3 \cdot 1 = 3 \]

### Step 5: Calculate the total favorable outcomes for either event A or event B.
Since events A and B are mutually exclusive (they cannot happen at the same time), we can add their probabilities:

\[ \text{Total favorable outcomes} = \text{Favorable outcomes for A} + \text{Favorable outcomes for B} = 3 + 3 = 6 \]

### Step 6: Calculate the probability of selecting either event A or event B.
The probability \(P(A \cup B)\) is given by:

\[ P(A \cup B) = \frac{\text{Total favorable outcomes}}{\text{Total ways}} = \frac{6}{15} \]

### Step 7: Simplify the probability.
\[ P(A \cup B) = \frac{6}{15} = \frac{2}{5} \]

### Final Answer:
The probability that 3 women and 1 man or 1 woman and 3 men are selected is \(\frac{2}{5}\).

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