Question

Three coins are tossed simultaneously 100 times with the following frequencies of different outcomes: Outcome No head One head Two heads Three heads Frequency 14 38 36 12 If the three coins are simultaneously tossed again, compute the probability of: (i) 2 heads coming up (ii) 3 heads coming up (iii) At least one head coming up (iv) Getting more heads than tails (v) Getting more tails than heads Select the correct answer from above options

Answer 1

We know, Probability of an event = (Number of Favorable outcomes) / (Total Numbers of outcomes) In this case, total numbers of outcomes = 100. (i) Probability of 2 Heads coming up = \(\frac{36}{100}\) = 0.36 (ii) Probability of 3 Heads coming up = \(\frac{12}{100}\) = 0.12 (iii) Probability of at least one head coming up = \(\frac{(38+36+12)}{100}\) = \(\frac{86}{100}\) = 0.86 (iv) Probability of getting more Heads than Tails = \(\frac{(36+12)}{100}\) = \(\frac{48}{100}\) = 0.48 (v) Probability of getting more tails than heads = \(\frac{(14+38)}{100 }\) = \(\frac{52}{100}\) = 0.52