Probability - Playing a Game with Two Fair Spinners

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Jack plays a game with two fair spinners, A and B, where he wins if one spinner lands on an odd number and the other on an even number. Spinner A has odd numbers 3, 5, and 7, while Spinner B has even numbers 2, 4, and 6. The probability of winning in a single round is calculated as 0.55 by adding the probabilities of the two winning scenarios. When playing twice, the probabilities remain independent, so the overall probability of winning both times is the product of the single round probabilities. The final conclusion is that the probability of winning both games is 0.55 squared, which equals approximately 0.3025.
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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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