# Each round Mike and Dean toss coin each. Mike tosses a not fair coin in which the probability to get heads is 0.6. Dean tosses a not fair coin in which the probability to get heads is 0.1. they toss the coins till they get the same results at the same time. What is the probability that there will be at most 5 rounds?

Each round Mike and Dean toss coin each. Mike tosses a not fair coin in which the probability to get heads is 0.6. Dean tosses a not fair coin in which the probability to get heads is 0.1. they toss the coins till they get the same results at the same time. What is the probability that there will be at most 5 rounds?
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Ramiro Sosa
Step 1
The game ends when Mike and Dean toss different results.
Mike: Heads $=0.6+Tails=0.4$
Dean: Heads $=0.1+Tails=0.9$
The probability of round being:
Both Heads $=0.60\cdot 0.40=0.24$
Both Tails $=0.10\cdot 0.90=0.09$
Both Same $=0.24+0.09=0.33$
Both Different $=1-0.33=0.67$
Step 2
What is the probability that of getting to round X?
$1=100\mathrm{%}$ (first round is always played)
$2=0.33$ (both same)
$3={0.33}^{2}=0.1089$ (both same, twice)
$4={0.33}^{3}=0.0359$ (both same, 3x)
$5={0.33}^{4}=0.0118$ (both same, 4x)
$6={0.33}^{5}=0.0039$ (both same, 5x)
What is the probability that there will be at most 5 rounds?
This is simply 1 minus the probability of getting to round 6
$\left(1-{0.33}^{5}\right)=0.99608$
###### Did you like this example?
Winston Todd
Step 1
The probability till Mike and Dean get the same results at most 5 rounds:
The final results can be...

The probability that Mike and Dean get the same results in a round $=0.6×0.1+\left(1-0.6\right)×\left(1-0.1\right)=0.42$.
Step 2
The probability that Mike and Dean do not get the same results in a round $=1-0.42=0.58$
Let X be the number of rounds until Mike and Dean get the same results. $X\sim Geo\left(0.42\right)$.
$P\left(X\le 5\right)=P\left(X=1\right)+P\left(X=2\right)+P\left(X=3\right)+P\left(X=4\right)+P\left(X=5\right)$