We are flipping a coin 6 times. What is the probability of the outcomes to be :
(Binomial distribution).
a) two tails.
b) At least 5 tails.
c) The most 2 heads
Solution:
"p=q=\\frac12"
"n=6"
"X\\sim Bin(n,p)"
a) P(two tails)=P(X=2)"=^6C_2(0.5)^2(0.5)^4=0.234375"
b) At least 5 tails)="P(X\\ge5)=P(X=5)+P(X=6)"
"=^6C_5(0.5)^5(0.5)^1+^6C_6(0.5)^6(0.5)^0\n\\\\=6(0.5)^6+(0.5)^6\n\\\\=0.109375"
c) P(The most 2 heads)"=P(X\\le2)=P(X=0)+P(X=1)+P(X=2)"
"=^6C_0(0.5)^0(0.5)^6+^6C_1(0.5)^1(0.5)^5+^6C_2(0.5)^2(0.5)^4\n\\\\=(0.5)^6+6(0.5)^6+15(0.5)^6\n\\\\=0.34375"
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