A fair coin is tossed 7 times. What is the probability that:
a) Exactly 5 heads appear?
b) At least two heads appear?
c) At most 4 heads appear?
and if you could explain how you got the answer, that would be great. thanks. (:
a) Exactly 5 heads appear?
b) At least two heads appear?
c) At most 4 heads appear?
and if you could explain how you got the answer, that would be great. thanks. (:

If it is a fair twosided coin flipped through a 3dimensional space, then p(heads) = 0.5
p(5 heads) = # ways to arrange 5 heads and 2 tails in a string of 7 flips
multiplied by the probability of 5 heads and then 2 tails
= (7 choose 5) * .5^5 * .5^2
= 7 x 6 / 2 * (1/2)^7
= 0.1640625
b)
1  P(0 heads)  P(1 heads) = P(>=2 heads)
P(0 heads) = 0.5^7
P(1 heads) = 7 x (0.5^6)(0.6)
1  8*(1/2)^7 = 0.9375
c)
P(0 heads) + P(1 heads) + P(2 heads) + P(3 heads) + P(4 heads)
They all have 0.5^7
1 + 7 + (6*7/2) + (6*7*5)/6 + (4*5*6*7)/(1*2*3*4)
1 + 7 + 21 + 35 + 35
99 (0.5)^7 = 0.7734375
p(5 heads) = # ways to arrange 5 heads and 2 tails in a string of 7 flips
multiplied by the probability of 5 heads and then 2 tails
= (7 choose 5) * .5^5 * .5^2
= 7 x 6 / 2 * (1/2)^7
= 0.1640625
b)
1  P(0 heads)  P(1 heads) = P(>=2 heads)
P(0 heads) = 0.5^7
P(1 heads) = 7 x (0.5^6)(0.6)
1  8*(1/2)^7 = 0.9375
c)
P(0 heads) + P(1 heads) + P(2 heads) + P(3 heads) + P(4 heads)
They all have 0.5^7
1 + 7 + (6*7/2) + (6*7*5)/6 + (4*5*6*7)/(1*2*3*4)
1 + 7 + 21 + 35 + 35
99 (0.5)^7 = 0.7734375