In a bombing of the Nathula pass, the Indian troops have to destroy a bridge on the pass. The bridge is such that it is destroyed when exactly 2 bombs hit it. A MIG-27 is dispatched in order to do the bombs hit it. A MIG-27 is dispatched in order to do the bombing. Flt Lt. Rakesh Sharma needs to ensure that there is at least 97% probability for the bridge to be destroyed. He knows that when he drops a bomb on the bridge the probability of the bomb hitting the bridge is 90%. Weather conditions and visibility being poor he is unable to see the bridge from his plane. How many bombs does he need to drop to be 95% sure that the bridge will be destroyed?

Options

A. 3
B. 4
C. 5
D. 6

Answer

To solve the problem step by step, we need to calculate the probability of exactly 2 bombs hitting the bridge when a certain number of bombs are dropped. The probability of hitting the bridge with a single bomb is given as 0.9, and the probability of missing is 0.1. We want to find the minimum number of bombs (n) that need to be dropped so that the probability of exactly 2 hits is at least 97%.

### Step 1: Define the Variables
- Let \( p = 0.9 \) (the probability of hitting the bridge).
- Let \( q = 1 - p = 0.1 \) (the probability of missing the bridge).
- Let \( n \) be the number of bombs dropped.

### Step 2: Use the Binomial Probability Formula
The probability of getting exactly \( k \) successes (hits) in \( n \) trials (bombs dropped) is given by the binomial probability formula:
\[ P(X = k) = \binom{n}{k} p^k q^{n-k} \]
where \( \binom{n}{k} \) is the binomial coefficient.

### Step 3: Set Up the Equation for Exactly 2 Hits
We want to find \( P(X = 2) \) for different values of \( n \):
\[ P(X = 2) = \binom{n}{2} (0.9)^2 (0.1)^{n-2} \]

### Step 4: Calculate for Different Values of \( n \)
We will calculate \( P(X = 2) \) for \( n = 3, 4, 5, \) and \( 6 \).

#### For \( n = 3 \):
\[ P(X = 2) = \binom{3}{2} (0.9)^2 (0.1)^{3-2} = 3 \times (0.81) \times (0.1) = 0.243 \]

#### For \( n = 4 \):
\[ P(X = 2) = \binom{4}{2} (0.9)^2 (0.1)^{4-2} = 6 \times (0.81) \times (0.01) = 0.0486 \]

#### For \( n = 5 \):
\[ P(X = 2) = \binom{5}{2} (0.9)^2 (0.1)^{5-2} = 10 \times (0.81) \times (0.001) = 0.0081 \]

#### For \( n = 6 \):
\[ P(X = 2) = \binom{6}{2} (0.9)^2 (0.1)^{6-2} = 15 \times (0.81) \times (0.0001) = 0.0001215 \]

### Step 5: Calculate the Total Probability
To find the total probability of at least 2 hits, we need to sum the probabilities of exactly 2 hits and more than 2 hits. We can calculate \( P(X \geq 2) \) by using the complement:
\[ P(X \geq 2) = 1 - P(X < 2) = 1 - (P(X = 0) + P(X = 1)) \]

### Step 6: Check Which \( n \) Gives at Least 97%
After calculating the probabilities for \( n = 3, 4, 5, \) and \( 6 \), we find:
- For \( n = 3 \): \( P(X = 2) = 0.243 \) (not sufficient)
- For \( n = 4 \): \( P(X = 2) = 0.486 \) (not sufficient)
- For \( n = 5 \): \( P(X = 2) = 0.081 \) (not sufficient)
- For \( n = 6 \): \( P(X = 2) = 0.01215 \) (not sufficient)

### Conclusion
After checking the probabilities, we find that dropping 3 bombs gives us a probability of hitting the bridge that is less than 97%. Therefore, we need to drop more bombs.

### Final Answer
To ensure at least a 97% probability of destroying the bridge, Flt Lt. Rakesh Sharma needs to drop **6 bombs**.