The probability that a candidate secure a seat in Engineering through EAMCET is `1/10`Seven candidate are selected at random from a centre.The probability that exactly two will get seats is

Options

A. `15(0.1)^2(0.9)^5`
B. `20(0.1)^2(0.9)^5`
C. `21(0.1)^2(0.9)^5`
D. `23(0.1)^2(0.9)^5`

Answer

To solve the problem, we will use the binomial probability formula. The steps are as follows:

### Step 1: Identify the parameters
We are given:
- The probability of success (a candidate securing a seat) \( p = \frac{1}{10} = 0.1 \)
- The probability of failure (a candidate not securing a seat) \( q = 1 - p = 1 - 0.1 = 0.9 \)
- The number of trials (candidates selected) \( n = 7 \)
- The number of successes we want (candidates securing seats) \( r = 2 \)

### Step 2: Write the binomial probability formula
The binomial probability formula is given by:
\[ P(X = r) = \binom{n}{r} p^r q^{n-r} \]
where:
- \( \binom{n}{r} \) is the binomial coefficient, calculated as \( \frac{n!}{r!(n-r)!} \)

### Step 3: Calculate the binomial coefficient
We need to calculate \( \binom{7}{2} \):
\[ \binom{7}{2} = \frac{7!}{2!(7-2)!} = \frac{7 \times 6}{2 \times 1} = 21 \]

### Step 4: Substitute the values into the formula
Now we substitute \( n = 7 \), \( r = 2 \), \( p = 0.1 \), and \( q = 0.9 \) into the formula:
\[ P(X = 2) = \binom{7}{2} (0.1)^2 (0.9)^{7-2} \]
\[ P(X = 2) = 21 \times (0.1)^2 \times (0.9)^5 \]

### Step 5: Calculate the probability
Now we calculate \( (0.1)^2 \) and \( (0.9)^5 \):
\[ (0.1)^2 = 0.01 \]
\[ (0.9)^5 = 0.59049 \]
Now substituting these values back:
\[ P(X = 2) = 21 \times 0.01 \times 0.59049 \]
\[ P(X = 2) = 21 \times 0.01 \times 0.59049 = 0.1239 \]

### Step 6: Final result
Thus, the probability that exactly two candidates will secure seats is approximately \( 0.1239 \).