Answer:
You need to throw the pair of dice 11 times for you to get a probability of at least .85 of getting a sum of 7 at least once.
Step-by-step explanation:
From the information given:
a pair of dice consists of two dice.
if a pair of dice is thrown, the probability of getting 7 Pr(7) = [tex]\dfrac{6}{36}[/tex]
Pr (7) = [tex]\dfrac{1}{6}[/tex]
Now, to estimate the probability of getting a sum of 7 at least once, we have:
Pr'(7)' = 1 - Probability of getting not sum is 7 ( 0.85 )
Pr'(7)' = 1 - 0.85
Pr'(7)' = 0.15
Probability of getting not sum is 7 = [tex]^nC_o \times (\dfrac{1}{6})^0 \times ( 1 - \dfrac{1}{6})^n = 0.15[/tex]
[tex]1 \times (\dfrac{5}{6})^n = 0.15[/tex]
[tex](\dfrac{5}{6})^n = 0.15[/tex]
[tex]n \times In \ \dfrac{5}{6} = In (0.15)[/tex]
n × ( - 0.18232) = -1.89711
n = [tex]\dfrac{-1.89711}{-0.18232}[/tex]
n = [tex]10.405[/tex]
n [tex]\simeq[/tex] 11
You need to throw the pair of dice 11 times for you to get a probability of at least .85 of getting a sum of 7 at least once.
