A room has three lamp sockets. From a collection of 10 light bulbs of which only six ar good, three bulbs are selected at random and placed in the sockets. What is the probability that there will be light in the room ?

Answer

To solve the problem of finding the probability that there will be light in the room when three bulbs are selected from a collection of 10 bulbs (6 good and 4 defective), we can follow these steps:

### Step 1: Determine Total Combinations of Selecting 3 Bulbs
The total number of ways to select 3 bulbs from 10 is given by the combination formula:

\[ \text{Total ways} = \binom{10}{3} \]

Calculating this:

\[ \binom{10}{3} = \frac{10!}{3!(10-3)!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 \]

### Step 2: Determine Combinations of Selecting 3 Defective Bulbs
Next, we find the number of ways to select 3 defective bulbs from the 4 defective bulbs:

\[ \text{Ways to select 3 defective bulbs} = \binom{4}{3} \]

Calculating this:

\[ \binom{4}{3} = \frac{4!}{3!(4-3)!} = \frac{4}{1} = 4 \]

### Step 3: Calculate the Probability of Selecting 3 Defective Bulbs
The probability of selecting 3 defective bulbs (which means there will be no light in the room) is given by the ratio of the number of ways to select 3 defective bulbs to the total ways to select 3 bulbs:

\[ P(\text{no light}) = \frac{\text{Ways to select 3 defective}}{\text{Total ways}} = \frac{4}{120} = \frac{1}{30} \]

### Step 4: Calculate the Probability of Having Light in the Room
The probability of having light in the room (which means at least one bulb is good) is the complement of the probability of having no light:

\[ P(\text{light}) = 1 - P(\text{no light}) = 1 - \frac{1}{30} = \frac{29}{30} \]

### Final Answer
Thus, the probability that there will be light in the room is:

\[ \boxed{\frac{29}{30}} \]

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