概率文字題

1.
到E地有兩種方法：
i. A→B→E
ii. A→B→C→E

P(A→B→E) = 1*1/3 = 1/3
P(A→B→C→E) = 1*1/3*1/2 = 1/6

P(到達E) = 1/3 + 1/6 = 1/2


2.
P(乘電車沒有遲到) = 2/7*(1- 2/5) = 2/7 * 3/5 = 6/35
P(乘巴士沒有遲到) = (1- 2/7)*(1- 1/5) = 5/7 * 4/5 = 4/7
P(沒有遲到) = 6/35 + 4/7 = 26/35

希望幫到你！^^

1)A man travel from A to B

Case 1
The man go directly to E
=1/3

Case 2
The man go to C, then E
=1/3×1/2
=1/6

Case 3,4,5
The man go to D, then H
or
The man go to D, then G
or
The man go to C, then F

As the man didn't arrive E
=0

Probability of the man arrive E
=1/3+1/6
=1/2

2)He either travel by tram or bus

Case 1
He travel by tram and didn't late for school
=2/7×(1-2/5)
=6/35

Case 2
He travel by bus and didn't late for school
=(1-2/7)×(1-1/5)
=20/35

Combining two cases:
=6/35+20/35
=26/35

1)
這人能到達E地的概率
= 到達B概率* 選擇E概率(E,C,D) + 到達B概率* 選擇C概率*選擇E概率
= 1* 1/3 + 1 * 1/3*1/2
= 1/3 + 1/6
=1/2

2)
他在某天上學沒有遲到的概率
= 乘搭電車的概率 * 乘搭電車沒有遲到 + 乘搭巴士的概率 * 乘搭巴士沒有遲到
= 2/7 * (1-2/5) + (1-2/7) * (1 - 1/5)
= 2/7 * 3/5 + 5/7 * 4/5
= 6/35 + 20 / 35
= 26/35