A game is played by rolling a six sided die which has four red faces and two blue faces. One turn co

icedagecs 2022-07-09 Answered
A game is played by rolling a six sided die which has four red faces and two blue faces. One turn consists of throwing the die repeatedly until a blue face is on top or the die has been thrown 4 times
Adnan and Beryl each have one turn. Find the probability that Adnan throws the die more turns than Beryl
I tried : Adnan throws two times and Beryl throws once = 2 3 x 1 3
Adnan throws three times and Beryl throws once = 4 9 x 1 2
Adnan throws three times and Beryl throws twice = 4 9 x 2 3
Adnan throws four times and Beryl throws once = 8 27 x 1 2
Adnan throws four times and Beryl throws twice = 8 27 x 2 3
Adnan throws four times and Beryl throws three times = 8 27 x 4 9
The answer says 0.365
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Answers (2)

Caiden Barrett
Answered 2022-07-10 Author has 20 answers
When Beryl throws the die once, Adnan can throw it 2,3 or 4 times. The required probability is
1 3 Beryl=1 ( 1 1 3 Adnan=1 )
Similarly, when Beryl throws the die 2 times, Adnan may throw 3 or 4 times, giving the required probability
2 3 1 3 Beryl=2 ( 1 ( 1 3 Adnan=1 + 2 3 1 3 Adnan=2 ) )
and when Beryl throws it 3 times, Adnan throws the die 4 times. The last throw could result in a red face or a blue face. So the probability of this case is
2 3 2 3 1 3 Beryl=3 ( 2 3 2 3 2 3 [ 2 3 + 1 3 ] Adnan=4 )
The sum of these terms yields the required answer.
Not exactly what you’re looking for?
Ask My Question
Ellen Chang
Answered 2022-07-11 Author has 5 answers
Well, A throws it more often than B in the following cases:
B throws 1, A throws 2,3,4
B throws 2, A throws 3,4
B throws 3, A throws 4

And now you go and check for the all-together probability of these events. For example:B throws 1, A throws 2,3,4:
P ( B = 1 ) = 2 6 and P ( A { 2 , 3 , 4 } ) = P ( A 1 ) = 4 6 . And finally P ( B = 1 A 1 ) = P ( B = 1 ) P ( A 1 ) = 2 9 . Note that I used that the events are independent.
Not exactly what you’re looking for?
Ask My Question

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

You might be interested in

asked 2022-06-22
Machine 1 is currently working. Machine 2 will be put in use at time t from now. If the lifetime of machine i is exponential with rate λ i , i = 1 , 2 , what is the probability that machine 1 is the first machine to fail?

The solution is as follows:
P ( M 1 < M 2 ) = P ( M 1 < M 2 | M 1 < t ) P ( M 1 < t ) + P ( M 1 < M 2 | M 1 > t ) P ( M 1 > t ) .
Now, I understand this solution. It states, in light of Bayes's Rule*, that we desire the probability that the lifetime of machine 1 is less than the lifetime of machine 2 and the lifetime of machine 1 is less than t in addition to the probability that the lifetime of machine 1 is less than the lifetime of machine 2 and the lifetime of machine 1 is greater than t. That all makes sense to me. My question, however, is why is it not simply
P ( M 1 < M 2 ) = P ( M 1 < t ) + P ( M 1 < M 2 | M 1 > t ) .
Linguistically, this seems to satisfy the solution to the problem; i.e., "the probability that the lifetime of machine 1 is less than t, in addition to the probability that the lifetime of machine 1 is less than that of machine 2 given that the lifetime of machine 1 is greater than t. I understand that the first term here is actually the same as the solution's, and that the solution formulates it as it does for illustrative purposes; but the second term certainly has a different value than that of the solution's.

I hope my question here makes sense. If any further clarification is needed, please let me know. Also, I understand that this exact problem appears elsewhere on stackexchange, but it was more about the computation.

*The Bayes's Rule to which I'm referring is, of course, the following formulation:
P ( A B ) = P ( A | B ) P ( B )
asked 2022-06-15
You are told that:
1. P ( P ) = 0.15
2. P ( T | P ) = 0.91
3. P ( T | ¬ P ) = 0.04
4. P and T are not independent.

Whats P ( T )?

I've been really struggling with this one. I've tried substituting a bunch of stuff into the multiplication rule
P ( A B ) = P ( A ) P ( B | A )
and the addition rule
P ( A B ) = P ( A ) + P ( B ) P ( A B ) .
I've also tried drawing a conditional probability tree to help me see what's going on, but I just don't seem to be getting anywhere.
asked 2022-06-03
This example is from the book first course in probability Example 4a.
Let X denote a random variable that takes on any of the values −1, 0, and 1 with respective probabilities P { X = 1 } = .2 P { X = 0 } = .5 > P { X = 1 } = .3.
Solution.Let Y = X 2 . Then the probability mass function of Y is given by
P { Y = 1 } = P { X = 1 } + P { X = 1 } = .5
P { Y = 0 } = P { X = 0 } = .5
What rule is used to compute P { Y = 1 }? When Y = X 2 Why addition is used ?
asked 2022-07-23
When do you add probabilities?
asked 2022-04-30
I know that there is the addition rule of probability, but I want to understand the intuition behind it. Specifically, why does OR signifies addition in probability theory?
asked 2022-05-28
In Texas holdem, one is dealt a Decent Hand (Any pocket pair or any two broadway cards) 15 percent of the time. If there are three people left in the hand, I can use the probability addition rule to say at least one of those three people left will show up with a Decent Hand 45 percent of the time, correct? What about when there are 10 people left? Does someone show up with a Decent Hand 150% of the time?
Edit: Why does the addition rule not suited for this case?
asked 2022-08-19
Could we use the special addition rule for determining the probability that for one draw from a deck of cards, that the card is either a Queen or a Heart? Why or why not?

New questions