Question

help appreciated!!!!!c) Suppose one pig die was rolled. Find the probability that the roll was NOT a snouter

Answer 1

a) We can calculate the empirical probability of each outcome as the absolute frequency of each outcome divided by the total number of rolls.

Then, for "ide with no dot" wwe can calculate the probability as:

`p_1=(1344)/(3939)\approx0.34`

This is equal to the relative frequency of this class.

We can calculate the others as:

`\begin{gathered} p_2=(1294)/(3939)\approx0.33 \\ \\ p_3=(767)/(3939)\approx0.19 \\ \\ p_4=(365)/(3939)\approx0.09 \\ \\ p_5=(137)/(3939)\approx0.03 \\ \\ p_6=(32)/(3939)\approx0.01 \end{gathered}`

We then can complete the table as:

b) Now we have to calculate the probability of a pig landing on its side with no dot OR on its side with a dot.

This includes the first two categories in the table.

As we are calculating the probability that any of the two happens, we can add the probabilities to find the probabilities of the event:

`P=p_1+p_2=0.34+0.33=0.67`

c) In this case, we have to calculate the probability of the roll not being a snouter.

This can be think of the cmplement of the evnent "snouter" and its probability is equal to 1 minus the probability of rolling a snouter.

This is because it includes all the other possible events but the snouter.

Then, we can calculate it as:

`\begin{gathered} P=1-p_5 \\ P=1-0.03 \\ P=0.97 \end{gathered}`

Answer:

a) Table pasted in the answer

b) The probability is 0.67

c) The probability is 0.97