Question

6A dice is tossed 60 times and the number it lands on isrecorded.Number 1 2 35Frequency7 12 10 6 15 10After being tossed 180 times how many times would youexpect it to land on:a) 4b) >3c) Evenw

Answer 1

Ok the first thing we have to do is calculate the approximate values of each number's probability. That probability tells how often will the dice land on a certain number. To approximate it you have to divide every frequency by the total amount of times the dice was tossed. Let's do that for every number:

$\begin{gathered} P(1)=\frac{7}{60}=0.118 \\ P(2)=\frac{12}{60}=0.2 \\ P(3)=\frac{10}{60}=0.166 \\ P(4)=\frac{6}{60}=0.1 \\ P(5)=\frac{15}{60}=0.25 \\ P(6)=\frac{10}{60}=0.166 \end{gathered}$

Now we can solve the question. For example if you want to know how many times the dice is expected to land on 4 you just need to multiply the probability for that number, which I called P(4), for the total amount of times the dice was tossed, which are 180 times:

$P(4)\cdot180=0.1\cdot180=18$

And that's the answer for a.

For item b we need to make more operations. We are being asked how many times the dice is expected to land on a number greater than 3. For this we'll have to use the probability for all the numbers on the dice that are greater than 3: 4, 5 and 6. We'll need to add all these probabilities and multiply the result od that sum by 180:

$(P(4)+P(5)+P(6))\cdot180=(0.1+0.25+0.166)\cdot180=0.516\cdot180=92.88$

Since 92.88 isn't a whole number we round it to 93 and that's the answer for item b.

We can repeat our calculations for item c but using the probabilities of the even numbers instead. Said numbers are 2, 4 and 6:

$(P(2)+P(4)+P(6))\cdot180=(0.2+0.1+0.166)\cdot180=0.466\cdot180=83.88$

Rounding 83.88 we have 84, the solution to item c.