Assuming all the coins are fair coins with equal probability =0.5
p
=
0.5
of heads or tails, we can model the probability of X out of N coins landing head side up with the binomial distribution:
(=∣)=()(1−)−
P
(
x
=
X
∣
p
)
=
(
N
X
)
p
X
(
1
−
p
)
N
−
X
Substituting the probability =0.5
p
=
0.5
and =4
N
=
4
into this pdf, we get:
(=)=(4)0.50.54−
P
(
x
=
X
)
=
(
4
X
)
0.5
X
0.5
4
−
X
To get the probability that three or more heads are obtained, we sum all the probabilities for ≥3
x
≥
3
, i.e.
(≥3)=∑=34(4)0.50.54−
P
(
x
≥
3
)
=
∑
x
=
3
4
(
4
X
)
0.5
X
0.5
4
−
X
(≥3)=(43)0.54+(44)0.54
P
(
x
≥
3
)
=
(
4
3
)
0.5
4
+
(
4
4
)
0.5
4
(≥3)=4(0.54)+0.54
P
(
x
≥
3
)
=
4
(
0.5
4
)
+
0.5
4
(≥3)=5(0.54)
P
(
x
≥
3
)
=
5
(
0.5
4
)
(≥3)=5(116)
P
(
x
≥
3
)
=
5
(
1
16
)
(≥3)=516=0.3125
P
(
x
≥
3
)
=
5
16
=
0.3125
Note: This particular question is simple enough to do without the above equation. One must recognize that there are 24
2
4
possible outcomes in the sample space of this four coin flip. Four of these outcomes have one coin which landed tails (one outcome for each different value of coin) i.e. three heads, and only one outcome exists where all coins are heads. These make up 5 of the 16 possible outcomes. Each are equally likely due to the equal probability of heads or tails for each coin, so we get the same answer as above.