A room has 3 lamps . From a collection of 10 light bulbs of which 6 are no good , a person selects 3 at random and puts them in a socket . What is the probability , that he will have light ?

Answer

To solve the problem, we need to find the probability that at least one of the selected light bulbs is good. Here's a step-by-step solution:

### Step 1: Understand the Problem
We have a total of 10 light bulbs, out of which 6 are defective (no good) and 4 are good. We need to select 3 bulbs at random and determine the probability that at least one of them is good.

### Step 2: Calculate the Total Ways to Choose 3 Bulbs
The total number of ways to choose 3 bulbs from 10 is given by the combination formula \( \binom{n}{r} \), where \( n \) is the total number of items, and \( r \) is the number of items to choose.

\[ \text{Total ways} = \binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 \]

### Step 3: Calculate the Ways to Choose 3 Defective Bulbs
Next, we calculate the number of ways to choose 3 defective bulbs from the 6 defective ones.

\[ \text{Ways to choose 3 defective bulbs} = \binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \]

### Step 4: Calculate the Probability of Choosing 3 Defective Bulbs
The probability of choosing 3 defective bulbs (which means no light) is the ratio of the number of ways to choose 3 defective bulbs to the total ways to choose 3 bulbs.

\[ P(\text{3 defective}) = \frac{\text{Ways to choose 3 defective}}{\text{Total ways}} = \frac{20}{120} = \frac{1}{6} \]

### Step 5: Calculate the Probability of Having Light
The probability of having at least one good bulb is the complement of the probability of having all defective bulbs.

\[ P(\text{at least 1 good}) = 1 - P(\text{3 defective}) = 1 - \frac{1}{6} = \frac{5}{6} \]

### Final Answer
Thus, the probability that the person will have light is:

\[ \boxed{\frac{5}{6}} \]