GCE A Level H2 Maths Paper 2
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7. A computer game simulates a tennis match between two players, A and B. The match consists of at most three sets. Each set is won by either A or B, and the match is won by the first player to win two sets.
The simulation uses the following rules.
- The probability that A wins the first set is 0.6.
- For each set after the first, the conditional probability that A wins that set, given that A won the preceding set, is 0.7.
- For each set after the first, the conditional probability that B wins that set, given that B won the preceding set, is 0.8.
Calculate the probability that
(i) A wins the second set, [2]
(ii) A wins the match, [3]
(iii) B won the first set, given that A wins the match. [3]
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0.42? 0.512? 0.109?
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so fast finish already ah?
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hannor
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i dun like to do probability
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I also don't like..... Statistics lai worse.
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so dun like, must train myself do more, then can tuition pple
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Can give exact steps?
I think I wrote those answers tamago suggested.
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i) 0.6×0.7 = 0.42
ii) P(AA)+P(ABA)+(BAA)
= 0.6×0.7+0.6×0.3×0.2+0.4×0.2×0.7
= 0.512
iii) 0.4×0.2×0.7 ÷ 0.512 = 0.109 -
i) should be 0.5 cos u missed out the part where A loses 1st and wins 2nd. 0.6x0.7 + 0.4x0.2= 0.5
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is it? i thought it's a cond prob divide by prob? :?
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lol. dunno lei. actually i also :S. but i heard drawing the tree diagram will get that answer. :/
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- The probability that A wins the first set is 0.6.
- For each set after the first, the conditional probability that A wins that set, given that A won the preceding set, is 0.7.
Let A1, A2, ... = A win first set, 2nd set, etc.
P(A2|A1) = P(A2^A1) / P(A1) = 0.7
P(A2^A1) = 0.7P(A1) = 0.42
Then how to find P(A2)? Or is P(A2^A1) already the answer to (i)?
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Gosh. i just typed a chunk and they tell me i cant post until 20 mins is up. :/
What if A loses first and then wins the 2nd? you are only finding A wins 1st and A wins 2nd leh.
10) A group of diplomats is to be chosen to represent 3 islands, K, L and M. The group is to consist of displomats and is chosen from a set of 12 displomats consisting of 3 from K, 4 from L and 5 from M. Find the no. of ways in which the group can be chosen if it includes:i) 2 diplomats from K, 3 from L and 3 from M
ii) diplomats from L and M only
iii) at least 4 diplomats from M
iv) at least 1 diplomat from each island. -
Originally posted by X1ao-lang:
lol. dunno lei. actually i also :S. but i heard drawing the tree diagram will get that answer. :/
I see 2 marks, then the rest 3 marks, then probabiltity not my strongest so I just quickly hamtam the 2 marks first.
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haha. nvm lah. 98/100. still zai. wahaha. wait. are you taking ur As this year??!?!?!
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yeah. i put 120, 9, 175, 7560 for that.
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oh wow. coolio. 7560!!!! too many lah. how can. how to get 175 ah.
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I got 120, 9, 210 and 7560
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the first 2 quite give away. i got 120, 9, 210 and 485.
7560 is... more than the number of ways they can choose their committee already. of 12 choose 8 = 400+ -
Originally posted by ^tamago^:
yeah.
i put 120, 9, 175, 7560 for that.
i got 7560 too, but it looked really big -
got stuck at the table tennis question too
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Why is 10(iv) 485?
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Is it : 12C8 - (no diplomats from K + no diplomats from L + no diplomats from M) = 12C8 - ( 9C8 + 8C8 + 0 ) = 495 - (9+1) = 485
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Originally posted by ^tamago^:
yeah.
i put 120, 9, 175, 7560 for that.
tennis qsn part 3 i wrong. dang.but i can tell u tt ur ans for part iv) of the diplomat qsn is wrong.
there is no permutation, only combination, cos here order is not important.
and taboo got the correct explanation.