Thread: Probability Problems

a) Cards numbered 1–10 are dealt out in pairs (without replacement). The sum of one person’s pair is 10. What is the probability that both cards contain odd numbers?


b) A soccer team’s star player plays for the full game in 50% of matches, part of the game in 40% and takes no part in the other 10%. When he plays for the full game the probability of a win for his team is 0.7; when he plays for part of the game it is 0.5; when he doesn’t play it is 0.3.
Draw a tree diagram for this data.
Hence calculate (i) the probability of a win for the team; (ii) the probability that he had played for the full match, given that the team won.

Well for part a I got 2/5. The numbers I got that added up to ten were 3-7 and 1-9. that's 4 out of ten possible numbers, reduced and got 2/5's....I''m assumin no replacement means that 5/5 isn't an option in which case it'd be 3/5......I hope I've helped.....

Originally Posted by mr_motivator

a) Cards numbered 1–10 are dealt out in pairs (without replacement). The sum of one person’s pair is 10. What is the probability that both cards contain odd numbers?


[snip]

The given condition is that the pair adds up to 10. So the restricted sample space is (1, 9), (2, 8), (3, 7), (4, 6), (6, 4), (7, 3), (8, 2), (9, 1).

Four of these have cards that are both odd. Therefore ....

Originally Posted by Itachi888Uchiha

Well for part a I got 2/5. The numbers I got that added up to ten were 3-7 and 1-9. that's 4 out of ten possible numbers, reduced and got 2/5's....I''m assumin no replacement means that 5/5 isn't an option in which case it'd be 3/5......I hope I've helped.....

Sorry but this is not correct.