Probability - Playing a Game with Two Fair Spinners

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Homework Help Overview

The problem involves calculating the probability of winning a game played with two fair spinners, where one spinner must land on an odd number and the other on an even number. The spinners have specific sets of numbers they can land on, and the game is played twice.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of a probability tree diagram to calculate winning probabilities and question the multiplication of probabilities for multiple rounds. There are differing values presented for the probability of winning in a single round, leading to confusion about the calculations.

Discussion Status

Participants are actively engaging with the problem, questioning assumptions about the independence of spins and the calculations involved. Some guidance has been offered regarding the correct formulation of the winning probability, but there is no clear consensus on the final probability value.

Contextual Notes

There seems to be uncertainty regarding the impact of previous spins on subsequent spins, as well as discrepancies in the calculated probabilities for winning in a single round.

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Homework Statement


Jack plays a game with two fair spinners, A and B. Spinner A can land on the number 2 or 3 or 5 or 7

Spinner B can land on the number 2 or 3 or 4 or 5 or 6

Jack spins both spinners.
He wins the game if one spinner lands on an odd number and the other spinner lands on an even number.

Jack plays the game twice.
Work out the probability that Jack wins the game both times

3. The Attempt at a Solution

If you do a probability tree diagram...

I get (1/4 x 2/5) + (3/4x3/5)^3 = 0.191125 That's for the first time he plays.
But then playing twice keeps the same probability, right? If not, why?
 
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Natasha1 said:

Homework Statement


Jack plays a game with two fair spinners, A and B. Spinner A can land on the number 2 or 3 or 5 or 7

Spinner B can land on the number 2 or 3 or 4 or 5 or 6

Jack spins both spinners.
He wins the game if one spinner lands on an odd number and the other spinner lands on an even number.

Jack plays the game twice.
Work out the probability that Jack wins the game both times

3. The Attempt at a Solution


If you do a probability tree diagram...

I get (1/4 x 2/5) + (3/4x3/5)^3 = 0.191125 That's for the first time he plays.
But then playing twice keeps the same probability, right? If not, why?

I get a completely different value for the probability of winning in a single round. I don't know why you are multiplying probabilities (as in (3/4 × 3/5)^3).

As to the second question: if he wins on the first round, will that affect the spinners in any way? Would the situation look exactly the same as it did before the first spin?
 
Is it just (1/4 x 2/5) x (3/4x3/5) = 0.045
 
Natasha1 said:
Is it just (1/4 x 2/5) x (3/4x3/5) = 0.045
No.
P(win) = P(S1 even & S2 odd) + P(S1 odd & S2 even).
 
Ray Vickson said:
No.
P(win) = P(S1 even & S2 odd) + P(S1 odd & S2 even).

Ah so (1/4 x 2/5) + (3/4x3/5) = 0.55
 
Natasha1 said:
Ah so (1/4 x 2/5) + (3/4x3/5) = 0.55
Yes.
 
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Much appreciated Ray, thanks so much :)
 

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