P(0)=P(3) = (1/2)^3
P(1)=P(2) = 3*(1/2)^3
Because 3C1 = 3C2 = 3
If we flip a coin three times what is the theoretical probability of getting zero heads, one heads, two heads, and three heads PLEASE ANSWER
2 answers
p heads = p tails = 0.5
lets call p heads favorable = p = .5
p tails unfavorable = q = 1-p = .5
P r times in n trials = C(n,r) p^r q^(n-r) where C(n,r) = n!/[r!(n-r)! defined = 1 if r = 0
so here n = 3
Zero heads
n = 3, r = 0
P = 1 * .5^0 * .5^(3) = .5^3 = .125 or 1/8
by the way that is the same as zero taails = 3 heads :)
now the hard one
Two heads, one tail
n = 3, r = 2
C(3,2) = 3!/ [ 2!(1!)] = 3
P = 3 * .5^2 * .5^1 = 3* .125 = 3/8
lets call p heads favorable = p = .5
p tails unfavorable = q = 1-p = .5
P r times in n trials = C(n,r) p^r q^(n-r) where C(n,r) = n!/[r!(n-r)! defined = 1 if r = 0
so here n = 3
Zero heads
n = 3, r = 0
P = 1 * .5^0 * .5^(3) = .5^3 = .125 or 1/8
by the way that is the same as zero taails = 3 heads :)
now the hard one
Two heads, one tail
n = 3, r = 2
C(3,2) = 3!/ [ 2!(1!)] = 3
P = 3 * .5^2 * .5^1 = 3* .125 = 3/8
82 answers