Three coins are tossed together. Find the probability of getting:
(i) At least two heads
(ii) At most two heads
(iii) No head
Step-by-Step Solution
Key Concept: $S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$ (8 outcomes).<br>(i) At least 2 heads: $\{HHH, HHT, HTH, THH\} \Rightarrow 4 \Rightarrow P = 4/8 = 1/2$.<br>(ii) At most 2 heads: all except $\{HHH\} \Rightarrow 7 \Rightarrow P = 7/8$.<br>(iii) No head: $\{TTT\} \Rightarrow 1 \Rightarrow P = 1/8$.
(i) $P(\text{At least 2 Heads}) = 4/8 = 1/2$. [1.0 Mark]
(ii) $P(\text{At most 2 Heads}) = 7/8$. [1.0 Mark]
(iii) $P(\text{No Head}) = 1/8$. [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Part (i) $P(\ge 2 \text{ Heads}) = 1/2$: 1.0 Mark
Part (ii) $P(\le 2 \text{ Heads}) = 7/8$: 1.0 Mark
Part (iii) $P(\text{No Head}) = 1/8$: 1.0 Mark
Correct Answer: