Question

A game of chance consist of spinning an arrow which comes to rest pointing at one of the numbers 15,20,25,30,35,40,45,50 and these are equally likely outcomes. What is the probability that it will point at
(i) an even number? (ii) at a number multiple of 5
(iii) at a number divisible by 3 (iv) at a prime number.

Answer 1

The sample is
`S={15,20,25,30,35,40,45,50}:n(S)=8`
(i) Let A be the event that arrow points at an even number.

Then `A={20,30,40,50} :.n(A)=4`
`:.P(A)=(n(A))/(n(S))P(A)=4/8=1/2`
(ii) Let B be the event that the arrow points at a number multiple of 5.
Then `B={15,20,25,30,35,40,45,50} :. n(B)=8`
`P(B)=(n(B))/(n(S)) :.P(B)=8/8=1`
(iii) Let C be the event that the arrow points at a number divisible by 3.
Then `C={15,30,45} :. n(C)=3`
`P(C)=(n(C))/(n(S)) :.P(C)=3/8`
(iv) Let D be the even the arrow points at a prime number.
Then `D={ }, :.n(D)=0`
`:.P(D)=(n(D))/(n(S)):.P(D)=0/8=0`
Ans (i) `1/2` (ii) 1 (iii) `3/8` (iv) 0