Ben and Allison each decide to wager 1 unit against the other person on flips of an unfair coin, with probability 0.6 of landing head, until one of them runs out of money. When the flip lands on head, Ben wins 1 unit from Allison; and when the coin lands on tail, Allison wins 1 from Ben. At the start of the contest, Ben has 30 units and Allison has 45 units. Find (a) the average number of flips needed until Ben is eventually broke, (b) the average number of flips needed until Allison is eventually broke, and (c) the average number of flips needed until either Ben or Allison is eventually broke.

(A) 75 flips

(B) 75 flips

(C) 75 flips

Step-by-step explanation:

Normally, the coin will have to land on tail 30 sure times (Allison will win 30 units from Ben) before Ben will be broke.

This does not mean that the first 30 flips will favour Allison (that Ben will get broke in the first 30 flips) hence we use the probability of tail to ascertain how many times the coin will be flipped until 30 tails are gotten. And for each sure tail, the coin will be tossed/flipped a number of times before the probability of tail will be one.

Also note that the coin used for this experiment is a biased or unfair coin. That is, both sides - head and tail - don't have an equal probability of occurrence which is 1/2 or 0.5

Same method applies for Head. The coin will have to land on Head 45 sure times (where in each time, probability of head occurring is 1) before Allison will be broke.

(A) Average number of flips needed until Ben is broke

This is equal to the average number of flips it takes for a tail to be gotten for sure X the number of units Ben will lose until he's broke.

The number of flips it takes to get a tail for sure is 2.5 flips.

How? 3 flips will get p(T) = 1.2

Hence 2.5 flips will get p(T) = 1

And that is when Ben loses a unit of his money

Now, if in 2.5 flips we have a tail, it will take X flips for us to get 30 tails

Cross multiply:

X = (2.5×30) / 1 = 75 flips

(B) Using the same cross multiplication method, we find that it takes 1.667 flips for Ben to win 1 unit of Allison's money because in 2 flips, the probability of head is 1.2, which is above 1.

So since it takes 1.667 for Allison to lose 1 unit of his money, it will take X flips for Allison to lose 45 units of his money

Cross multiply:

X = (45×1.667)/1 = 75 flips

(C) We would have had to use a formula for this one but there is no need, since it takes the same 75 flips for either Ben or Allison to get broke.

a is the correct

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)