A bag contains 3 black balls and 4 white balls. Two balls are drawn from the bag without replacement. Find the probability that both balls are black given that one of them is black.
A. 1/7
B. 1/6
C. 1/5
D. 1/4
p.s. plz do not just give me the ans... calculate it with steps
thanks for yr help!
f5 maths mc
2009-02-26 2:12 am
回答 (3)
2009-02-26 2:37 am
✔ 最佳答案
P(not all balls are white) = 1 - 4/7*3/6 = 5/7 =P(at least one ball is black)
P(all balls are black) = 3/7*2/6
= 1/7
So P(E) = 1/7 over 5/7
= 1/5
Answer is C.
2009-02-25 18:53:08 補充：
method 2 :
『B1B2 B1B3 B2B3』 are require cases , 3 cases
W1B1 W2B1 W3B1 W4B1
W1B2 W2B2 W3B2 W4B2
W1B3 W2B3 W3B3 W4B3 are possible but not require cases,12cases
So P(E) = 3/(3+12) = 1/5
2009-02-26 3:46 am
A：2白球事件，B：2黑球事件，C：至少1黑球事件
P(A)=(4/7)*(3/6)=12/42
P(B)=(3/7)*(2/6)=6/42
P(C)=1－P(A)=30/42
P(B|C)=P(BC)/P(C)=P(B)/PC)=(6/42)/(30/42)=1/5
2009-02-26 2:43 am
3over7 times 2over6
= 3 x 2
--- ----
7 6
= 1
----
7
= a
= 3 x 2
--- ----
7 6
= 1
----
7
= a
收錄日期: 2021-04-21 22:02:35
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090225000051KK01092