Question

your turn
calculate all the theoretical probabilities for the given binomial experiment. then draw a histogram of
the probability distribution and use it to find the specified probabilities.

1. a binomial experiment consists of flipping a weighted coin for 4 trials where getting heads is considered a

success. the coin lands heads up 40% of the time. find the probability of a) getting 2 or more heads and b)
the probability of getting fewer than 4 heads.

number of
heads
0
1
2
3
4
theoretical
probability

Explanation:

Step1: Calculate theoretical probabilities

The binomial probability formula is \(P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}\), where \(n = 4\), \(p=0.4\), and \(C(n,k)=\frac{n!}{k!(n - k)!}\)

• For \(k = 0\):

\(C(4,0)=\frac{4!}{0!(4-0)!}=1\)
\(P(X = 0)=1\times(0.4)^{0}\times(0.6)^{4}=0.1296\)

• For \(k = 1\):

\(C(4,1)=\frac{4!}{1!(4 - 1)!}=4\)
\(P(X = 1)=4\times(0.4)^{1}\times(0.6)^{3}=4\times0.4\times0.216 = 0.3456\)

• For \(k = 2\):

\(C(4,2)=\frac{4!}{2!(4-2)!}=6\)
\(P(X = 2)=6\times(0.4)^{2}\times(0.6)^{2}=6\times0.16\times0.36=0.3456\)

• For \(k = 3\):

\(C(4,3)=\frac{4!}{3!(4 - 3)!}=4\)
\(P(X = 3)=4\times(0.4)^{3}\times(0.6)^{1}=4\times0.064\times0.6 = 0.1536\)

• For \(k = 4\):

\(C(4,4)=\frac{4!}{4!(4-4)!}=1\)
\(P(X = 4)=1\times(0.4)^{4}\times(0.6)^{0}=0.0256\)

Step2: Calculate probability of getting 2 or more heads

\(P(X\geq2)=P(X = 2)+P(X = 3)+P(X = 4)\)
\(P(X\geq2)=0.3456+0.1536 + 0.0256=0.5248\)

Step3: Calculate probability of getting fewer than 4 heads

\(P(X\lt4)=1 - P(X = 4)\)
\(P(X\lt4)=1-0.0256 = 0.9744\)

Answer:

• Theoretical Probabilities: \(0.1296\) (for \(0\) heads), \(0.3456\) (for \(1\) head), \(0.3456\) (for \(2\) heads), \(0.1536\) (for \(3\) heads), \(0.0256\) (for \(4\) heads)
• a) \(P(X\geq2)=0.5248\)
• b) \(P(X\lt4)=0.9744\)