Question

Find the probanility of getting.

a.fewer than 4 head.

b.exactly 4 head.

C.not more than 4 head.

d.two or more heads.

Answer 1

Solution:

Let we flip coin 4 times. First we need to find probability of 0-head, 1-head, 2-head, 3-head and 4-head.

Number of possible outcomes: $N=2^{n};$ here $n$ - number of flips.

Number of ways: $^{n}C_{m}=\frac{n!}{(n-m)!m!};$ here $m$ - number of heads. So,

$^{4}C_{0}=\frac{4!}{(4-0)!0!}=1;$

$^{4}C_{1}=\frac{4!}{(4-1)!1!}=4;$

$^{4}C_{2}=\frac{4!}{(4-2)!2!}=6;$

$^{4}C_{3}=\frac{4!}{(4-3)!3!}=4;$

$^{4}C_{4}=\frac{4!}{(4-4)!4!}=1;$

$N=2^{4}=16;$

$P(0)=\frac{1}{16}=0.0625$ - probability of 0 head;

$P(1)=\frac{4}{16}=0.25$ - probability of 1 head;

$P(2)=\frac{6}{16}=0.375$ - probability of 2 heads;

$P(3)=\frac{4}{16}=0.25$ - probability of 3 heads;

$P(4)=\frac{1}{16}=0.0625$ - probability of 4 heads.

a) Fewer than 4 heads:

$P(m<4)=P(0)+P(1)+P(2)+P(3)=0.0625+0.25+0.375+0.25=0.9375;$

b) Exactly 4 heads:

$P(m=4)=P(4)=0.0625;$

c) Not more than 4 heads:

$P(m<=4)=P(0)+P(1)+P(2)+P(3)+P(4)=0.0625+0.25+0.375+0.25+0.0625=1;$

d) Two or more heads:

$P(m>=2)= P(2)+P(3)+P(4)=0.375+0.25+0.0625=0.6875.$

Answer:

a) $P(m<4)=93.75\%;$

b) $P(m=4)=6.25\%;$

c) $P(m<=4)=100\%;$

d) $P(m>=2)=68.75\%;$