Question

In a game of chance ,the spinning arrow rests at one of the numbers 1, 2, 3 , 4, 5, 6, 7, 8. All these are equally likely outcomes. Find the probabilities of the following events.
A)The arrow rests at an odd no.
B)It rests at a prime number.
C)It rests at a multiple of 2

Answer 1

• We need to find the probability of the given events.

• Now since, all of these outcomes are equally likely, the probability of the arrow to rest on any of the number = 1/8

Now,

A) Let E be the event for arrow to be resting on an odd number.                                                           The possible outcomes are 1, 3, 5, 7.

So, P(E)= number of favorable outcomes/total number of outcomes

P(E)= 4/8 = 1/2

B) Let A be the event describing arrow resting on a prime number.                         The favorable outcomes are 2, 3, 5, 7.

So, P(A) = 4/8 = 1/2

C) Let B be the event describing arrow to rest on a multiple of 2.                                The favorable outcomes are 2, 4, 6, 8.

So, P(E) = 4/8 = 1/2.

Answer 2

Answer:

a.)  the probability that the arrow rests on an odd number  =  1 / 2

b.) probability that the arrow rests at a prime number   = 1 / 2

c.) probability that the arrow rests at a multiple of two  = 1 / 2

Step-by-step explanation:

a.) odd numbers are: 1, 3, 5, 7

   the probability that the arrow rests on an odd number  = $\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{1}{2}$

b.) prime numbers are 2,3,5,7

    probability that the arrow rests at a prime number   = $\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{1}{2}$

c.) multiples of 2  are 2, 4, 6, 8

 probability that the arrow rests at a multiple of two  = $\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{1}{2}$