Answer:
4 heads
Explanation:
(X=k)=(
k
n
)⋅p
k
⋅(1−p)
n−k
In this case,
�
=
10
n=10,
�
=
4
k=4, and
�
=
0.5
p=0.5 because the coin is fair.
Probability of exactly 4 heads:
�
(
�
=
4
)
=
(
10
4
)
⋅
(
0.5
)
4
⋅
(
0.5
)
10
−
4
P(X=4)=(
4
10
)⋅(0.5)
4
⋅(0.5)
10−4
Now, calculate this probability:
�
(
�
=
4
)
=
(
10
4
)
⋅
(
0.5
)
4
⋅
(
0.5
)
6
P(X=4)=(
4
10
)⋅(0.5)
4
⋅(0.5)
6
You can calculate
(
10
4
)
(
4
10
) as 210.
So,
�
(
�
=
4
)
=
210
⋅
(
0.5
)
4
⋅
(
0.5
)
6
P(X=4)=210⋅(0.5)
4
⋅(0.5)
6
Simplify,
�
(
�
=
4
)
=
210
⋅
0.0625
⋅
0.015625
P(X=4)=210⋅0.0625⋅0.015625
Now, calculate the product:
�
(
�
=
4
)
=
0.20508
P(X=4)=0.20508
So, the probability of getting exactly 4 heads is approximately 0.2051.
Probability of at least 4 heads:
To find the probability of getting at least 4 heads, you need to consider all possibilities of getting 4, 5, 6, 7, 8, 9, or 10 heads and sum up their probabilities.
�
(
�
≥
4
)
=
�
(
�
=
4
)
+
�
(
�
=
5
)
+
�
(
�
=
6
)
+
�
(
�
=
7
)
+
�
(
�
=
8
)
+
�
(
�
=
9
)
+
�
(
�
=
10
)
P(X≥4)=P(X=4)+P(X=5)+P(X=6)+P(X=7)+P(X=8)+P(X=9)+P(X=10)
You've already calculated
�
(
�
=
4
)
P(X=4) in the previous part. Now, you can calculate the probabilities for
�
(
�
=
5
)
P(X=5),
�
(
�
=
6
)
P(X=6),
�
(
�
=
7
)
P(X=7),
�
(
�
=
8
)
P(X=8),
�
(
�
=
9
)
P(X=9), and
�
(
�
=
10
)
P(X=10) using the same formula as before, changing
�
k accordingly.
Sum up these probabilities to find
�
(
�
≥
4
)
P(X≥4).
Remember that in each case,
�
=
0.5
p=0.5 and
�
=
10
n=10.
Once you've calculated and summed up these probabilities, you will get the probability of getting at least 4 heads.