Question

If a fair coin is tossed 10 times. Find the probability of getting at most six heads.

Answer 1

Probability of getting at most 6 heads

= 1 – [P (7 Heads) + P(8 Heads) + P(9 Heads) + P(10 Heads)]

= 1 – [¹⁰C₇ (0.5)¹⁰ + ¹⁰C₈ (0.5)¹⁰ + ¹⁰C₉ (0.5)¹⁰ + ¹⁰C₁₀ (0.5)¹⁰]  ......`[∵ p = 1/2, q = 1/2]`

= `1 - [(10 xx 9 xx 8)/(3 xx 2) + (10 xx 9)/2 + 10 + 1] (0.5)^10`

= 1 –[120 + 45 + 11] (0.5)¹⁰

= `1 - 176 xx (1/2)^10`

= `(1024 - 176)/1024`

= `848/1024`

= `53/64`