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