Question

a dice can have 6 sides and are numbered 1,2,3,4,5, and 6. The odds of getting an odd number are the same. The chance of getting an even number is the same. The chance of getting an odd number is twice the chance of getting an even number.
a. Determine the opportunity to get the number 3.
b. The dice is thrown three times. Determine the odds of getting two numbers 5 and one number 4.
c. The dice is tossed a hundred times. Use the approximation to determine the chances of getting an even number at most 37 times.

Answer 1

The chance of getting an odd number is twice the chance of getting an even number

So., say prob of getting even numbers = x

& prob of getting odd numbers = 2x

3 odd and 3 even numbers and total probability=1

Hence 3*x + 3*2x=1

3x+6x=1

x=1/9

So prob of getting even numbers i.e., 2,4,6 =1/9 each

& prob of getting odd numbers i.e., 1,3,5 =2/9 each

a) Determine the opportunity to get the number 3 = 2/9 as shown above

b)The dice is thrown three times. Determine the odds of getting two numbers 5 and one number 4

= 2/9*2/9*1/9 * 3 (2/9 is the prob of 5 and 1/9 is the prob of 4 but these numbers can come in 3 possible orders i.e. 5,5,4 and 5,4,5 and 4,5,5)

ans=2/9*2/9*1/9 * 3

= 0.01646091

c)The dice is tossed a hundred times. Use the approximation to determine the chances of getting an even number at most 37 times can be solved using binomial theorm

P(X=x) = nCx * p^x * q^(n-x)

x<=37 for at most 37

p=1/3 (1/9+1/9+1/9=1/3 prob of even nunmber)

q=2/3(prob of odd number)

n=100

now we have to put x=0 to 37 and sum up all the values and for that we need a calculator/software

Below I have answered the same in R

Ans=> 0.8123113

Hope the above answer has helped you in understanding the problem. Please upvote the ans if it has really helped you. Good Luck!!