Probability - Playing a Game with Two Fair Spinners

  • Thread starter Natasha1
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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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Answers and Replies

  • #2
Ray Vickson
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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?

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?
 
  • #3
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Is it just (1/4 x 2/5) x (3/4x3/5) = 0.045
 
  • #4
Ray Vickson
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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).
 
  • #5
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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
 
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Ray Vickson
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Much appreciated Ray, thanks so much :)
 

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