## thespanishscholar you win a game if you roll a die and get a 2 or 5 you play 60 times what is the probability you will win the game at least 40 times--- i know that is a cumulative probability i know that 60 *.333=19.98 i know that the answer for winning 15 out of the 60 times because the book includes some answers and that answer is 20--- i know when i do it on a binomial calculator that is at stattrek.com it gave me the cumulative as 1.32 also when i did it by the number of successes which in this case is 40 so i did 40*.333=13.32 so i am still confused as to which probability is the right one-- one year ago one year ago

• This Question is Open
1. thespanishscholar

i got my .333 when i did (1/6)+(1/6)

2. thespanishscholar

can someone tell me if i got the right probability in there somewhere

3. dumbcow

this is a binomial distribution problem http://en.wikipedia.org/wiki/Binomial_distribution#Probability_mass_function

4. thespanishscholar

thank you dumbcow! I appreciate the information

5. kropot72

The probability of rolling a 2 or 5 = 1/6 + 1/6 = 1/3 This can be solved by using the normal approximation of the binomial distribution with the mean and the standard deviation calculated as follows: $mean=np=60\times \frac{1}{3}=20$ $standard\ deviation=\sqrt{np(1-p)}=\sqrt{20\times \frac{2}{3}}=3.65$ The cumulative probability of the number of wins being 39 or fewer is found from a standard normal distribution table. The z-score for X = 39 is$z=\frac{39-20}{3.65}=5.2$ reference to a standard normal distribution table shows that the probability of the number of wins being 39 or fewer is 1.00. Therefore there is zero probability that you will win the game at least 40 times.

6. thespanishscholar

thank you kropot72-this is really helpful-

7. kropot72

You're welcome :) The normal approximation of the binomial distribution is very useful when the number of trials becomes large.