Probability

Question

A lottery game works as follows: The player draws a numbered ball at a random from an urn containing five balls numbered 1,2,3,4,and 5. If the number on the ball is even, the player loses the game and receives no points; if the number on the ball is odd, the player receives the number of points indicated on the ball. Afterwords, he or she replaces the ball in the urn and draws again. On each subsequent turn, the player loses the game if the total of the all the numbers drawn becomes even, and gets another turn( after receiving the number of points indicated on the ball and then replacing the ball in the urn) each time the total remains odd.
What is the probability that the player accumulates exactly 7 points and then loses on the next turn?

OA is [spoiler]198/3125[/spoiler]

Leon1984 · Answer

If he has 7 points already, the next ball will either be even (he looses) or odd (the sum becomes even and he looses) so any ball will make him loose, so it's 100%.

However, how can he get to 7 points? In order to get to seven, he has to get an odd ball and then an even ball, but he looses whenever the ball is even, so he can't really get to 7 points? He can also get an odd ball and then again an odd ball, but the sum becomes even so he looses? so is it 0% since it is impossible?

What am I missing?

Abdulla · Answer

Leon1984 wrote:If he has 7 points already, the next ball will either be even (he looses) or odd (the sum becomes even and he looses) so any ball will make him loose, so it's 100%.

However, how can he get to 7 points? In order to get to seven, he has to get an odd ball and then an even ball, but he looses whenever the ball is even, so he can't really get to 7 points? He can also get an odd ball and then again an odd ball, but the sum becomes even so he looses? so is it 0% since it is impossible?

What am I missing?

I think you're missing this point. At the first picked, If he picked a ball with an even number he loses, but If he picked a ball with an odd number he count the number and return the ball back and then he pick another one until the sum of the points that he got adds up to an even number.

ssuarezo · Answer

Abdulla wrote:Leon1984 wrote:If he has 7 points already, the next ball will either be even (he looses) or odd (the sum becomes even and he looses) so any ball will make him loose, so it's 100%. However, how can he get to 7 points? In order to get to seven, he has to get an odd ball and then an even ball, but he looses whenever the ball is even, so he can't really get to 7 points? He can also get an odd ball and then again an odd ball, but the sum becomes even so he looses? so is it 0% since it is impossible? What am I missing? I think you're missing this point. At the first picked, If he picked a ball with an even number he loses, but If he picked a ball with an odd number he count the number and return the ball back and then he pick another one until the sum of the points that he got adds up to an even number. I think it's too wordy and not clear question.

Harbinder · Answer

There are 5 ways in which this can happen

1+2+4+(any odd number) -> probability P1= (1/5)(1/5)(1/5)(3/5) since 2 and 4 can switch places multiply the probablity by 2 so P1 = 6/625

1+2+2+2+(any odd number) -> P2 = (3/3125)
1+3+2+2+(any odd number) -> P3 = (3/625)
1+3+4+(any odd number) -> P2 = (3/125)
1+5+2+(any odd number) -> P2 = (3/125)

Total Probability = P1+P2+P2+P3+P4+P5 = 198/3125

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