Question

In a game called taxation and evasion, a player rolls a pair of dice. if on any turn the sum is 7, 11, or 12, the player gets audited. otherwise, she avoids taxes. suppose a player takes 5 turns at rolling the dice. the probability that she gets audited no more than 2 times is

Answer 1

The probability that she gets audited no more than 2 times is 0.896482...

As, a pair of dice are rolling here, so the number of total possible outcome = (6×6) = 36

For Sum = 7 , the favorable outcomes are: (1,6) (2,5) (3,4) (4,3) (5,2) and (6,1)

For Sum = 11 , the favorable outcomes are: (5,6) and (6,5)

For Sum = 12 , the favorable outcome is: (6,6)

Probability = (Number of favorable outcomes)÷(Number of total outcomes)

So, $P(7)= \frac{6}{36} =\frac{1}{6}$

$P(11) = \frac{2}{36} = \frac{1}{18}$

$P(12) = \frac{1}{36}$

P( 7 or 11 or 12) = $\frac{1}{6}+\frac{1}{18}+\frac{1}{36} = \frac{6+2+1}{36}=\frac{9}{36} = \frac{1}{4}$

Here the total number of trials = 5 and the probability of getting audited = $\frac{1}{4}$

According the binomial distribution formula:

P(X) = (ⁿCₓ )(P)ˣ (1-P)ⁿ⁻ˣ

where P(X) is the probability of x successes out of n trials

Here, n= 5 and P = 1/4 and we need to find the probability of getting audited no more than 2 times. This means she can gets audited 0, 1 or 2 times.

So,

$P(X=0)+P(X=1) +P(X=2)\\\\= [^5C^0 (\frac{1}{4})^0 (1-\frac{1}{4})^5^-^0 ]+[^5C^1 (\frac{1}{4})^1 (\frac{3}{4})^5^-^1]+[^5C^2 (\frac{1}{4})^2 (\frac{3}{4})^5^-^2]\\\\= (\frac{3}{4})^5+ (5) (\frac{1}{4}) (\frac{3}{4})^4 +(10)(\frac{1}{4})^2 (\frac{3}{4})^3\\\\=0.237304... +0.395507...+0.263671...\\\\ = 0.896482...$

So, the probability that she gets audited no more than 2 times is 0.896482...