Two gamblers bet $1 each on the successive tosses of acoin. Each has a bank of $6. What is the probability that:a They break even after six tosses?b One play wins all the money after the 10th toss?

Part a: The probability of breaking even in 6 tosses is 0.3125.

Part b: The probability that one payer wins all the money after the 10th toss is 0.0264.

Explanation:

Part a

P(success)=1/2=0.5

P(Failure)=1/2=0.5

Now for the break-even at the sixth toss

P(Break Even)=P(3 success out of 6)

P(3 success out of 6)

So the probability of breaking even in 6 tosses is 0.3125.

Part b:

So the probability that one of the player wins all the money after the 10th toss is given as the tenth toss is given as a win so

Wins in 9 tosses is given as 9!/7!=72

The probability that the other person wins

Wins in 8 out of 10 tosses is given as 10!/8!(10-8)!=10!/8!2!=45

So the probability of all the money is won by one of the gambler after the 10th toss is given as

P=number of wins in 9 tosses-Number of wins in 10 tosses/total number of tosses

P=(72-45)/2^16

P=0.0264

So the probability that one payer wins all the money after the 10th toss is 0.0264.

the answer is d

brainliest! ?

i guess you could put raise the economic balance