Please explain the following for me.
1. two cards are drawn randomly from five cards A,B,C,D and E. Find the probability that card A is drawn while card C is not.
ans: 3/10
2. A bag contains 2 black balls, 2 green balls and 2 yellow balls. Peter repeats drawing one ball at a time randomly from the bag without replacement until a green ball is drawn. Find the probability that he needs at most 4 draws.
ans: 14/15
f.5 probability~please help!
2007-02-20 8:09 am
回答 (2)
2007-02-20 8:26 am
✔ 最佳答案
1.P(A and NOT C)
=P(A) x P(NOT C | A) + P(NOT C and NOT A) x P(A | NOT C and NOT A)
=1/5 x 3/4 + 3/5 x 1/4
=3/10
2.
P(A green with at most 4 draws)
=1 - P(A green with 5 draws)
=1 - 4/6 x 3/5 x 2/4 x 1/3
=14/15
2007-02-20 02:06:00 補充：
如果後面果到只係NOT C而唔NOT A即係第一張有機會抽到A就同前面果到重覆左了
2007-02-20 02:09:05 補充：
第二位直接x2因為無論A抽先定係A抽後兩者機率均相等例如有47個人排隊拿波仔無論你係第幾個抽抽中1號波既機會係1/47
2007-02-20 8:28 am
Q.1
Probability
= Prob (A is drawn in 1st place) x Prob (B, D or E is drawn in 2nd place) x 2
= 1/5 x 3/4 x 2
= 3/10
Q.2
Prob (first 4 draws are not green)
= 4/6 x 3/5 x 2/4 x 1/3
= 1/15
So,
prob (he needs at most 4 draws to get a green ball)
= 1 - 1/15
= 14/15.
Probability
= Prob (A is drawn in 1st place) x Prob (B, D or E is drawn in 2nd place) x 2
= 1/5 x 3/4 x 2
= 3/10
Q.2
Prob (first 4 draws are not green)
= 4/6 x 3/5 x 2/4 x 1/3
= 1/15
So,
prob (he needs at most 4 draws to get a green ball)
= 1 - 1/15
= 14/15.
收錄日期: 2021-04-12 21:22:23
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