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Finite Math Examples
x<1 , n=6 , p=5
Step 1
Subtract 5 from 1.
-4
Step 2
When the value of the number of successes x is given as an interval, then the probability of x is the sum of the probabilities of all possible x values between 0 and n. In this case, p(x<1)=P(x=0).
p(x<1)=P(x=0)
Step 3
Step 3.1
Use the formula for the probability of a binomial distribution to solve the problem.
p(x)=C06⋅px⋅qn-x
Step 3.2
Find the value of C06.
Step 3.2.1
Find the number of possible unordered combinations when r items are selected from n available items.
C06=Crn=n!(r)!(n-r)!
Step 3.2.2
Fill in the known values.
(6)!(0)!(6-0)!
Step 3.2.3
Simplify.
Step 3.2.3.1
Simplify the numerator.
Step 3.2.3.1.1
Expand (6)! to 6⋅5⋅4⋅3⋅2⋅1.
6⋅5⋅4⋅3⋅2⋅1(0)!(6-0)!
Step 3.2.3.1.2
Multiply 6⋅5⋅4⋅3⋅2⋅1.
Step 3.2.3.1.2.1
Multiply 6 by 5.
30⋅4⋅3⋅2⋅1(0)!(6-0)!
Step 3.2.3.1.2.2
Multiply 30 by 4.
120⋅3⋅2⋅1(0)!(6-0)!
Step 3.2.3.1.2.3
Multiply 120 by 3.
360⋅2⋅1(0)!(6-0)!
Step 3.2.3.1.2.4
Multiply 360 by 2.
720⋅1(0)!(6-0)!
Step 3.2.3.1.2.5
Multiply 720 by 1.
720(0)!(6-0)!
720(0)!(6-0)!
720(0)!(6-0)!
Step 3.2.3.2
Simplify the denominator.
Step 3.2.3.2.1
Expand (0)! to 1.
7201(6-0)!
Step 3.2.3.2.2
Subtract 0 from 6.
7201(6)!
Step 3.2.3.2.3
Expand (6)! to 6⋅5⋅4⋅3⋅2⋅1.
7201(6⋅5⋅4⋅3⋅2⋅1)
Step 3.2.3.2.4
Multiply 6⋅5⋅4⋅3⋅2⋅1.
Step 3.2.3.2.4.1
Multiply 6 by 5.
7201(30⋅4⋅3⋅2⋅1)
Step 3.2.3.2.4.2
Multiply 30 by 4.
7201(120⋅3⋅2⋅1)
Step 3.2.3.2.4.3
Multiply 120 by 3.
7201(360⋅2⋅1)
Step 3.2.3.2.4.4
Multiply 360 by 2.
7201(720⋅1)
Step 3.2.3.2.4.5
Multiply 720 by 1.
7201⋅720
7201⋅720
Step 3.2.3.2.5
Multiply 720 by 1.
720720
720720
Step 3.2.3.3
Divide 720 by 720.
1
1
1
Step 3.3
Fill the known values into the equation.
1⋅(5)0⋅(1-5)6-0
Step 3.4
Simplify the result.
Step 3.4.1
Multiply (5)0 by 1.
(5)0⋅(1-5)6-0
Step 3.4.2
Anything raised to 0 is 1.
1⋅(1-5)6-0
Step 3.4.3
Multiply (1-5)6-0 by 1.
(1-5)6-0
Step 3.4.4
Subtract 5 from 1.
(-4)6-0
Step 3.4.5
Subtract 0 from 6.
(-4)6
Step 3.4.6
Raise -4 to the power of 6.
4096
4096
4096