Answer:
[tex]\begin{array}{cccccc}x & {0} & {1} & {2} & {3} & {4} \ \\ P(x) & {0.0256} & {0.1536} & {0.3456} & {0.3456} & {0.1296} \ \end{array}[/tex]
Step-by-step explanation:
Given
[tex]p = 60\%[/tex]
[tex]n = 4[/tex]
Required
The distribution of x
The above is an illustration of binomial theorem where:
[tex]P(x) = ^nC_x * p^x *(1 - p)^{n-x}[/tex]
This gives:
[tex]P(x) = ^4C_x * (60\%)^x *(1 - 60\%)^{n-x}[/tex]
Express percentage as decimal
[tex]P(x) = ^4C_x * (0.60)^x *(1 - 0.60)^{n-x}[/tex]
[tex]P(x) = ^4C_x * (0.60)^x *(0.40)^{4-x}[/tex]
When x = 0, we have:
[tex]P(x=0) = ^4C_0 * (0.60)^0 *(0.40)^{4-0}[/tex]
[tex]P(x=0) = 1 * 1 *(0.40)^4[/tex]
[tex]P(x=0) = 0.0256[/tex]
When x = 1
[tex]P(x=1) = ^4C_1 * (0.60)^1 *(0.40)^{4-1}[/tex]
[tex]P(x=1) = 4 * (0.60) *(0.40)^3[/tex]
[tex]P(x=1) = 0.1536[/tex]
When x = 2
[tex]P(x=2) = ^4C_2 * (0.60)^2 *(0.40)^{4-2}[/tex]
[tex]P(x=2) = 6 * (0.60)^2 *(0.40)^2[/tex]
[tex]P(x=2) = 0.3456[/tex]
When x = 3
[tex]P(x=3) = ^4C_3 * (0.60)^3 *(0.40)^{4-3}[/tex]
[tex]P(x=3) = 4 * (0.60)^3 *(0.40)[/tex]
[tex]P(x=3) = 0.3456[/tex]
When x = 4
[tex]P(x=4) = ^4C_4 * (0.60)^4 *(0.40)^{4-4}[/tex]
[tex]P(x=4) = 1 * (0.60)^4 *(0.40)^0[/tex]
[tex]P(x=4) = 0.1296[/tex]
So, the probability distribution is:
[tex]\begin{array}{cccccc}x & {0} & {1} & {2} & {3} & {4} \ \\ P(x) & {0.0256} & {0.1536} & {0.3456} & {0.3456} & {0.1296} \ \end{array}[/tex]
