Subtopic: Discrete & Continuous Probability Distributions
Question Statement:
Consider an unfair coin. The probability of getting heads is 0.6. If you toss this coin twice,
what is the probability that the first or the second toss is heads?
Options:
(A) Q.56
(B) Q.64
(C) Q.84
(D) Q.96
Official Answer Key:
C
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE BT 2018 Q.11)**
**1. Governing Concept & Question Classification**:
- **Section**: BT
- **Question Type**: **MCQ** (1 Mark)
- **Topic**: Biomathematics / Probability Theory (Independent Events & Union)
**2. Detailed Step-by-Step Derivation**:
- Let $H_1$ be the event of getting heads on the first toss, and $H_2$ be the event of getting heads on the second toss.
- The coin is unfair with $P(H) = 0.6$. Consequently, $P(T) = 1 - 0.6 = 0.4$.
- The two tosses are independent events.
- We want to find the probability that the first OR the second toss is heads: $P(H_1 \cup H_2)$.
- **Method 1 (Complement rule)**:
The complement of "getting at least one head in two tosses" is "getting tails on both tosses" ($T_1 \cap T_2$):
$P( ext{Both Tails}) = P(T_1) imes P(T_2) = 0.4 imes 0.4 = 0.16$
$P(H_1 \cup H_2) = 1 - P( ext{Both Tails}) = 1 - 0.16 = 0.84$
- **Method 2 (Addition rule of probability)**:
$P(H_1 \cup H_2) = P(H_1) + P(H_2) - P(H_1 \cap H_2)$
$P(H_1 \cup H_2) = 0.6 + 0.6 - (0.6 imes 0.6) = 1.2 - 0.36 = 0.84$
- **Option-by-Option Evaluation**:
- **Option (A) [INCORRECT]**: 0.56 is $P( ext{exactly one head}) = 2 imes (0.6 imes 0.4) = 0.48$ or an arithmetic error.
- **Option (B) [INCORRECT]**: 0.64 is $1 - 0.36$, confusing probabilities.
- **Option (C) [CORRECT]**: 0.84 is the exact probability of obtaining heads on the first or second toss ($1 - 0.16 = 0.84$).
- **Option (D) [INCORRECT]**: 0.96 assumes an incorrect tail probability of 0.2.
**3. Final Answer / Verification Summary**:
- **Official GATE Answer Key**: **C**
- **Scoring Criteria**: +1.0 Mark (Negative marking: -0.33 marks).
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