Probability

1. Method #1 :
P(Peter needs at most 4 times)
= P(green at the 1st time) + P(green at the 2nd time) +
__P(green at the 3rd time) + P(green at the 4th time)
= 2/6 + (4/6)(2/5) + (4/6)(3/5)(2/4) + (4/6)(3/5)(2/4)(2/3)
= 1/3 + 4/15 + 1/5 + 2/15
= 14/15

Method #2 :
P(Peter needs at most 4 times)
= 1 - P(Peter needs more than 4 times)
= 1 - (4/6)(3/5)(2/4)(1/3)
= 1 - 1/15
= 14/15


2.
P(the rumour still has not returned to Alfred after it is told 10 times)
= 1 * 1 * (39/40)^8 (The first time is Alfred told the rumour and the
__second time is the one told by Alfred chose another classmate.)
= 0.816651803
= 0.817 (correct to 3 significant figures)


Take your Q#1 as example, finding the probability that Peter needs

"at most" 4 times means that you have to consider all cases less than

or equal to "4 times". On the other hand, finding the probability that

Peter needs "at least" 4 times means that you have to consider all

cases more than or equal to "4 times". The main point of "at least"

and "at most" is to include the boundary case.

2009-03-23 16:09:45 補充：
Correction for Q#2 :

P(the rumour still has not returned to Alfred after it is told 10 times)
= 1 * (39/40)^9 (The first time is Alfred told the rumour and so not a return.)

2009-03-23 16:09:54 補充：
(Then, the rumour spreads at random and Alfred could be told with a
probability of 1/40(A student cannot tell the rumour to himself.). So, in the
following 9 times, the rumour has a probability of 39/40 not returned to Alfred.)

= 0.796235508
= 0.796 (correct to 3 significant figures)

Q1: 13/15

Q2:0.796

Q1
P(he needs at most 4 draws)
=1-P(he needs at least 4 draws)
=1-(4/6x3/5x2/4x2/3)
=13/15


Q2
P(the rumour still has not returned to Alfred after it is told 10 times)
=40/40x(39/40)^9
=0.796

2009-03-18 16:11:30 補充：
at most 4 times means u can draw 1 time or 2 times or 3 times or 4 times,
but u cant draw 5 times

at least 4 times means that u can draw 4,5,6 and so forth which must be higher than or equal to 4