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
Question: 1
exactly one surgery is successful.
Show Hint
For small $n$, you can also think of the outcomes: $SFF, FSF, FFS$.
Each has probability $0.9 \times 0.1 \times 0.1 = 0.009$. Since there are 3 such outcomes, the total is $3 \times 0.009 = 0.027$.
Concept: • This is a case of independent trials following a Binomial distribution, where $n=3$.
• Probability of success $P(S) = p = 0.9$.
• Probability of failure $P(F) = q = 1 - p = 0.1$.
• The probability of exactly $k$ successes is given by $P(X=k) = \binom{n}{k} p^k q^{n-k}$.
Step 1: Identify the parameters
Number of patients (trials), $n = 3$.
Success rate, $p = 90\% = 0.9$.
Failure rate, $q = 1 - 0.9 = 0.1$.
Step 2: Set up the formula for exactly one success
We want to find $P(X=1)$:
\[ P(X=1) = \binom{3}{1} (p)^1 (q)^{3-1} \]
Step 3: Calculate the numerical value
\[ P(X=1) = 3 \times (0.9) \times (0.1)^2 \]
\[ P(X=1) = 3 \times 0.9 \times 0.01 \]
\[ P(X=1) = 0.027 \]
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Question: 2
From the same data as above, find the probability that at most two surgeries are successful.
Show Hint
"At most $n-1$" successes for $n$ trials is always $1 - P(\text{all successes})$.
This shortcut saves you from calculating and summing multiple individual binomial terms.
Concept: • "At most two" means $X \leq 2$, which includes $X=0, X=1,$ and $X=2$.
• It is easier to use the complement rule: $P(X \leq 2) = 1 - P(X=3)$.
Step 1: Calculate the probability of all three surgeries being successful
\[ P(X=3) = p^3 = (0.9)^3 \]
\[ P(X=3) = 0.9 \times 0.9 \times 0.9 = 0.729 \]
Step 2: Use the complement rule
The total probability is 1. The probability of at most two successes is:
\[ P(X \leq 2) = 1 - P(X=3) \]
\[ P(X \leq 2) = 1 - 0.729 \]
\[ P(X \leq 2) = 0.271 \]
