Q.59 If a fair coin is tossed two times, the probability that the first or the second toss will be heads is __________ (rounded off to two decimal places).

Q.59 If a fair coin is tossed two times, the probability that the first or the second toss
will be heads is __________ (rounded off to two decimal places).

When a fair coin is tossed two times, the probability that the first or second toss results in heads equals 0.75.

Sample Space

The sample space consists of four equally likely outcomes: HH, HT, TH, TT, each with probability \( \frac{1}{4} \).

\[ P(\text{first or second heads}) = \frac{3}{4} = 0.75 \]

Favorable outcomes where at least one toss shows heads are HH, HT, and TH.

Union Rule Calculation

Let \( E \) be heads on first toss (\( P(E) = \frac{1}{2} \), outcomes HH, HT) and \( F \) be heads on second toss (\( P(F) = \frac{1}{2} \), outcomes HH, TH).

The intersection \( E \cap F \) is HH (\( P(E \cap F) = \frac{1}{4} \)).

By inclusion-exclusion:

\[ P(E \cup F) = P(E) + P(F) – P(E \cap F) = \frac{1}{2} + \frac{1}{2} – \frac{1}{4} = \frac{3}{4} = 0.75 \]

Complementary Probability

The only unfavorable outcome is TT (both tails), with probability \( \frac{1}{4} \).

\[ P(\text{at least one head}) = 1 – P(\text{both tails}) = 1 – \frac{1}{4} = 0.75 \]

Common Errors

  • Adding probabilities directly (\( \frac{1}{2} + \frac{1}{2} = 1 \)) double-counts HH.
  • Assuming independence without subtraction overestimates by ignoring overlap.

This matches CSIR NET-style quantitative reasoning, emphasizing union rules and sample spaces.

 

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