Q.A survey was conducted on the patients who have undergone knee replacement surgeries. It was found that Robotic Knee replacement surgeries have a $90\%$ success rate. On a particular day, robotic surgery was performed on three patients, A, B and C, one after the other. Assuming that the success and failure of each surgery is independent of each other, find the probability that:

Concept: Independent Bernoulli trials with success probability $p = 0.9$ and failure probability $q = 0.1$.

Each surgery outcome is independent. We model this as three independent trials where success has probability $0.9$.

1. Exactly one surgery successful: We need exactly one success and two failures. The number of ways to choose which one surgery succeeds is $\binom{3}{1} = 3$. Probability = $3 \times (0.9)^1 \times (0.1)^2 = 3 \times 0.9 \times 0.01 = 0.027$.
2. At most two surgeries successful: "At most two" means 0, 1, or 2 successes. It's easier to use the complement: at most two successful = 1 − (all three successful). …

The problem gives us three independent trials (surgeries on A, B, and C), each with the same probability of success $p = 0.9$ and failure $q = 1 - p = 0.1$. This is a classic binomial setup, but with only three trials it's clearest to enumerate the outcomes directly rather than blindly plug into a formula.

Independence means the outcome of one surgery doesn't affect another. So the probability of any specific sequence—say, success for A, failure for B, success for C—is just the product of the individual probabilities.

(i) Exactly one surgery is successful

We need exactly one success and two failures. There are three ways this can happen, depending on which patient has the successful surgery:

1. A succeeds, B and C fail:

   $P(\text{S, F, F}) = 0.9 \times 0.1 \times 0.1 = 0.009$

2. B succeeds, A and C fail:

   $P(\text{F, S, F}) = 0.1 \times 0.9 \times 0.1 = 0.009$

3. C succeeds, A and B fail:

   $P(\text{F, F, S}) = 0.1 \times 0.1 \times 0.9 = 0.009$

These three outcomes are mutually exclusive (they can't happen simultaneously), so we add their probabilities:

Alternatively, using the binomial formula $\binom{n}{k} p^k q^{n-k}$ with $n=3$, $k=1$:

(ii) At most two surgeries are successful …

Method: Binomial probabilities for a few independent trials

With a small fixed number of independent success/failure trials, model the count as binomial and use the complement for "at most / at least."

Steps

Step 1: Fix the parameters.

Identify $n$ (number of trials), $p$ (success probability), $q=1-p$. The success count is $X\sim B(n,p)$.

Step 2: Exactly $k$ — include the arrangement count.

The $\binom{n}{k}$ factor counts which trials succeed; forgetting it is the classic error. …

Common Mistakes

Mistake 1: Dropping the arrangement factor for "exactly one success."

Why it's wrong: exactly one success among three can be A, B, or C, so multiply by $\binom{3}{1}=3$: $3(0.9)(0.1)^2=0.027$, not $0.009$. Correct approach: include the count of positions.

Mistake 2: Misreading "at most two" as "exactly two." …