Question

If a fair coin is tossed $10$ times find the probability of (a) exactly six head
(b) At least six heads
(c) at most six heads.

Answer 1

a) Probability of exactly six heads=${{10}_C}_6{\left(0.5\right)}^{10}$

b)Probability of at least 6 heads=

P(6 heads)+P(7 heads)+P(8 heads)+P(9 heads)+P(10 heads)

=${{10}_C}_6{\left(0.5\right)}^{10}+{{10}_C}_7{\left(0.5\right)}^{10}+{{10}_C}_8{\left(0.5\right)}^{10}+{{10}_C}_9{\left(0.5\right)}^{10}+{{10}_C}_{10}{\left(0.5\right)}^{10}$=386/1024

c))Probability of at most 6 heads=

1-P(7 heads)+P(8 heads)+P(9 heads)+P(10 heads)

=$1-{{10}_C}_7{\left(0.5\right)}^{10}+{{10}_C}_8{\left(0.5\right)}^{10}+{{10}_C}_9{\left(0.5\right)}^{10}+{{10}_C}_{10}{\left(0.5\right)}^{10}$=848/1024