Question

1. A carnival game offers a $100 cash prize for a game where the player tries to toss a ring onto one of many pegs. Alex will play the ring toss game five times, with an 8% chance of making any given throw.

   1. What is the probability that Alex tosses one of the five rings onto a peg?

   2. What is the probability that Alex tosses more than one of the five rings onto a peg?

   3. If Alex tossed five rings again and again, how many rings would land on a peg on average, that is, what's the mean number of successes he can expect?

Answer 1

Formula to be used : for binomial probability distribution, probability of getting x number of success (where p is the success probability) in n number of trials Is given by

P(X=x) = nCx * (p)x * (1 - p)n-x

a) Let X be the event that Alex tosses a ring onto a peg.

Number of throws = n = 5 ,

Success probability = p = 0.08

, Number of successes = x = 1

We will be using binomial probability distribution in this case .

So, probability that alex tosses one of the 5 rings onto a peg = P(X=1) = 5C1 (0.08)1 (0.92)4  = 0.286557 = 0.2866

b) P(X > 1 ) = 1 - P(X=0) - P(X=1) = 1 - 0.925  - 0.286557 = 0.05436

c) as per law of averages, long time average number of successes = expected number of successful throws = number of trials * success probability of each trial = n*p = 5*0.08 = 0.40

Here are the images in case you need them. .