Question:
Amar, Akbar, and Anthony write a test independently. The probability that Amar passes the test and Akbar fails the test is $\frac{3}{20}$. The probability that Akbar passes the test and Anthony fails the test is $\frac{1}{4}$. The probability that Amar and Anthony both pass the test is $\frac{2}{5}$. Then the probability that at least one of Amar, Akbar and Anthony fails the test is
Amar, Akbar, and Anthony write a test independently. The probability that Amar passes the test and Akbar fails the test is $\frac{3}{20}$. The probability that Akbar passes the test and Anthony fails the test is $\frac{1}{4}$. The probability that Amar and Anthony both pass the test is $\frac{2}{5}$. Then the probability that at least one of Amar, Akbar and Anthony fails the test is
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Using complementary probability is the fastest way to solve "at least one" problems.
Always check the validity of solved parameters (they must lie in $[0, 1]$) to eliminate extraneous algebraic solutions.
Always check the validity of solved parameters (they must lie in $[0, 1]$) to eliminate extraneous algebraic solutions.
Updated On: Jun 16, 2026
- $\frac{7}{10}$
- $\frac{3}{10}$
- $\frac{3}{5}$
- $\frac{4}{5}$
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question:
The problem involves three independent events (passing a test) with given compound probabilities.
We need to determine individual probabilities to find the probability of the complement event: "at least one of them fails".
Step 2: Key Formula or Approach:
For independent events $A, B, C$:
- $P(A \cap B') = P(A) \cdot P(B') = P(A)(1 - P(B))$
- $P(A \cap C) = P(A) \cdot P(C)$
- $P(\text{at least one fails}) = 1 - P(\text{all pass}) = 1 - P(A)P(B)P(C)$
Step 3: Detailed Explanation:
• Let $a$, $b$, and $c$ be the probabilities that Amar, Akbar, and Anthony pass the test, respectively.
• From the given conditions, we write the equations:
1) $a(1 - b) = \frac{3}{20}$
2) $b(1 - c) = \frac{1}{4}$
3) $ac = \frac{2}{5} \implies c = \frac{2}{5a}$
• Substitute $c$ into equation (2):
\[ b\left(1 - \frac{2}{5a}\right) = \frac{1}{4} \implies b\left(\frac{5a - 2}{5a}\right) = \frac{1}{4} \implies b = \frac{5a}{4(5a - 2)} \]
• Substitute this expression for $b$ into equation (1):
\[ a\left(1 - \frac{5a}{4(5a - 2)}\right) = \frac{3}{20} \]
\[ a\left(\frac{20a - 8 - 5a}{4(5a - 2)}\right) = \frac{3}{20} \]
\[ a\left(\frac{15a - 8}{4(5a - 2)}\right) = \frac{3}{20} \]
• Multiplying both sides by 20:
\[ 5a\left(\frac{15a - 8}{5a - 2}\right) = 3 \]
\[ 75a^2 - 40a = 15a - 6 \implies 75a^2 - 55a + 6 = 0 \]
• Factoring the quadratic:
\[ (5a - 3)(15a - 2) = 0 \implies a = \frac{3}{5} \text{ or } a = \frac{2}{15} \]
• Let us evaluate both cases:
Case 1: $a = \frac{3}{5}$
\[ c = \frac{2}{5\left(\frac{3}{5}\right)} = \frac{2}{3} \quad (\text{valid, since } 0 \le c \le 1) \]
\[ b = \frac{5\left(\frac{3}{5}\right)}{4\left(5\left(\frac{3}{5}\right) - 2\right)} = \frac{3}{4(3 - 2)} = \frac{3}{4} \quad (\text{valid}) \]
Case 2: $a = \frac{2}{15}$
\[ c = \frac{2}{5\left(\frac{2}{15}\right)} = 3 \quad (\text{invalid, since probability cannot exceed 1}) \]
• Thus, the individual probabilities are $a = \frac{3}{5}$, $b = \frac{3}{4}$, and $c = \frac{2}{3}$.
• Calculate the probability that all three pass:
\[ P(Amar \cap Akbar \cap Anthony) = a \cdot b \cdot c = \frac{3}{5} \cdot \frac{3}{4} \cdot \frac{2}{3} = \frac{3}{10} \]
• The probability that at least one fails is:
\[ P(\text{at least one fails}) = 1 - \frac{3}{10} = \frac{7}{10} \]
Step 4: Final Answer:
The probability that at least one fails the test is $\frac{7}{10}$.
The problem involves three independent events (passing a test) with given compound probabilities.
We need to determine individual probabilities to find the probability of the complement event: "at least one of them fails".
Step 2: Key Formula or Approach:
For independent events $A, B, C$:
- $P(A \cap B') = P(A) \cdot P(B') = P(A)(1 - P(B))$
- $P(A \cap C) = P(A) \cdot P(C)$
- $P(\text{at least one fails}) = 1 - P(\text{all pass}) = 1 - P(A)P(B)P(C)$
Step 3: Detailed Explanation:
• Let $a$, $b$, and $c$ be the probabilities that Amar, Akbar, and Anthony pass the test, respectively.
• From the given conditions, we write the equations:
1) $a(1 - b) = \frac{3}{20}$
2) $b(1 - c) = \frac{1}{4}$
3) $ac = \frac{2}{5} \implies c = \frac{2}{5a}$
• Substitute $c$ into equation (2):
\[ b\left(1 - \frac{2}{5a}\right) = \frac{1}{4} \implies b\left(\frac{5a - 2}{5a}\right) = \frac{1}{4} \implies b = \frac{5a}{4(5a - 2)} \]
• Substitute this expression for $b$ into equation (1):
\[ a\left(1 - \frac{5a}{4(5a - 2)}\right) = \frac{3}{20} \]
\[ a\left(\frac{20a - 8 - 5a}{4(5a - 2)}\right) = \frac{3}{20} \]
\[ a\left(\frac{15a - 8}{4(5a - 2)}\right) = \frac{3}{20} \]
• Multiplying both sides by 20:
\[ 5a\left(\frac{15a - 8}{5a - 2}\right) = 3 \]
\[ 75a^2 - 40a = 15a - 6 \implies 75a^2 - 55a + 6 = 0 \]
• Factoring the quadratic:
\[ (5a - 3)(15a - 2) = 0 \implies a = \frac{3}{5} \text{ or } a = \frac{2}{15} \]
• Let us evaluate both cases:
Case 1: $a = \frac{3}{5}$
\[ c = \frac{2}{5\left(\frac{3}{5}\right)} = \frac{2}{3} \quad (\text{valid, since } 0 \le c \le 1) \]
\[ b = \frac{5\left(\frac{3}{5}\right)}{4\left(5\left(\frac{3}{5}\right) - 2\right)} = \frac{3}{4(3 - 2)} = \frac{3}{4} \quad (\text{valid}) \]
Case 2: $a = \frac{2}{15}$
\[ c = \frac{2}{5\left(\frac{2}{15}\right)} = 3 \quad (\text{invalid, since probability cannot exceed 1}) \]
• Thus, the individual probabilities are $a = \frac{3}{5}$, $b = \frac{3}{4}$, and $c = \frac{2}{3}$.
• Calculate the probability that all three pass:
\[ P(Amar \cap Akbar \cap Anthony) = a \cdot b \cdot c = \frac{3}{5} \cdot \frac{3}{4} \cdot \frac{2}{3} = \frac{3}{10} \]
• The probability that at least one fails is:
\[ P(\text{at least one fails}) = 1 - \frac{3}{10} = \frac{7}{10} \]
Step 4: Final Answer:
The probability that at least one fails the test is $\frac{7}{10}$.
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