Question

Find the probability of exactly one 5 when a die is rolled 3 times 2) Find the probability of getting 3 heads when 8 coins are tossed. 3) A bag contains 4 red and 2 green balls. A ball is drawn and replaced 4 times. What is the probability of getting exactly 3 red balls and 1 green ball.

Answer 1

Find the probability of exactly one 5 when a die is rolled 3 times 

$P(1)$ Probability of exactly 1 successes

$\text{trials} = 3$

$p = \frac{1}{6}=0.167$ and $X = 1$

into a binomial probability distribution function (PDF). If doing this by hand,

apply the binomial probability formula:

$P(X) = \binom{n}{X} \cdot p^X \cdot (1-p)^{n-X}$

The binomial coefficient,$\binom{n}{X}$ is defined by

$\binom{n}{X} = \frac{n!}{X!(n-X)!}$

The full binomial probability formula with the binomial coefficient is

$P(X) = \frac{n!}{X!(n-X)!} \cdot p^X \cdot (1-p)^{n-X}$

where n is the number of trials, p is the probability of success on a single trial, and X

is the number of successes. Substituting in values for this problem, 

n=3

p=0.167

X=1

$P(1) = \frac{3!}{1!(3-1)!} \cdot 0.167^1 \cdot (1-0.167)^{3-1}$

Evaluating the expression, we have

P(1)=0.347638389

2) Find the probability of getting 3 heads when 8 coins are tossed.  

$P(3)$ Probability of exactly 3 successes

$\text{trials} = 8$

$p = \frac{3}{8}=0.375$ and $X = 3$

into a binomial probability distribution function (PDF). If doing this by hand,

apply the binomial probability formula:

$P(X) = \binom{n}{X} \cdot p^X \cdot (1-p)^{n-X}$

The binomial coefficient,$\binom{n}{X}$ is defined by

$\binom{n}{X} = \frac{n!}{X!(n-X)!}$

The full binomial probability formula with the binomial coefficient is

$P(X) = \frac{n!}{X!(n-X)!} \cdot p^X \cdot (1-p)^{n-X}$

where n is the number of trials, p is the probability of success on a single trial, and X

is the number of successes. Substituting in values for this problem, 

n=8

p=0.375

X=3

$P(3) = \frac{8!}{3!(8-3)!} \cdot 0.375^3 \cdot (1-0.375)^{8-3}$

Evaluating the expression, we have

P(3)=0.28163194656372

3) A bag contains 4 red and 2 green balls. A ball is drawn and replaced 4 times. What is the probability of getting exactly 3 red balls and 1 green ball.

we find probability for exact 1 green ball then automatically remaining balls are red.

$P(1)$ Probability of exactly 1 successes

$\text{trials} = 4$

$p = \frac{2}{6}=0.333$ and $X = 1$

into a binomial probability distribution function (PDF). If doing this by hand,

apply the binomial probability formula:

$P(X) = \binom{n}{X} \cdot p^X \cdot (1-p)^{n-X}$

The binomial coefficient,$\binom{n}{X}$ is defined by

$\binom{n}{X} = \frac{n!}{X!(n-X)!}$

The full binomial probability formula with the binomial coefficient is

$P(X) = \frac{n!}{X!(n-X)!} \cdot p^X \cdot (1-p)^{n-X}$

where n is the number of trials, p is the probability of success on a single trial, and X

is the number of successes. Substituting in values for this problem, 

n=4

p=0.333

X=1

$P(1) = \frac{4!}{1!(4-1)!} \cdot 0.333^1 \cdot (1-0.333)^{4-1}$

Evaluating the expression, we have

P(1)=0.395258962716