b*******2
发帖数: 3 | 1 if you have 100 different numbered cards, you pick 10 cards randomly every
time, and put them back afterwards. What's the probability of two cards
appearing together in two consecutive draws out of a five times | A**u
发帖数: 2458 | 2 试着解一下，
抽取5次，组成4个事件
E1,E2,E3,E4
Ei表示第i次抽取的10张卡，和第i+1次抽取的10张卡，至少有2张相同
P(Ei) = 1 - P(0张相同） - P (1张相同）
= 1 - 0.3304762 - 0.4079953
= 0.2615285
E1,E2,E3,E4至少发生一次的概率
P = 1 - 0.7384715 ** 4
= 0.2973959 | b*******2
发帖数: 3 | 3 that is correct, thanks. | r******g
发帖数: 13 | 4 How do you get P(1张相同）? just feel P (1张相同）< P(0张相同）, not sure
【在 A**u 的大作中提到】
: 试着解一下，
: 抽取5次，组成4个事件
: E1,E2,E3,E4
: Ei表示第i次抽取的10张卡，和第i+1次抽取的10张卡，至少有2张相同
: P(Ei) = 1 - P(0张相同） - P (1张相同）
: = 1 - 0.3304762 - 0.4079953
: = 0.2615285
: E1,E2,E3,E4至少发生一次的概率
: P = 1 - 0.7384715 ** 4
: = 0.2973959
| G*********e
发帖数: 614 | 5 P(0张相同) ＝combin(90,10)/combin(100,10)
P(1张相同) ＝10*combin(90,9)/combin(100,10)
【在 r******g 的大作中提到】
: How do you get P(1张相同）? just feel P (1张相同）< P(0张相同）, not sure
| D**********d
发帖数: 849 | 6 let q = \sum_{k = 0}^{9} \frac{90 - k}{100-k}, the probability of drawing a
card that drawn in the previous selection.
Then the probability of not consecutive draws is q^4. therefore the
probability of consecutive draws is 1-q^4. Note that it's 4, instead of 5. | c**********e
发帖数: 2007 | |
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