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Zolol [24]
4 years ago
15

Rewrite 24/5 as a mixed number in the lowest terms please help

Mathematics
1 answer:
tankabanditka [31]4 years ago
4 0
The answer is 4 4/5.
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A certain class has 20 students, and meets on Mondays and Wednesdays in a classroom with exactly 20 seats. In a certain week, ev
Olin [163]

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The probability that no one sits in the same seat on both days of that week is given by, P(\cap^{20}_{i=1}A_i^c)=\frac{1}{e}

Step-by-step explanation:

Given : A certain class has 20 students, and meets on Mondays and Wednesdays in a classroom with exactly 20 seats. In a certain week, everyone in the class attend both days. On both days, the students choose their seats completely randomly (with one student per seat).

To find : The probability that no one sits in the same seat on both days of that week ?

Solution :

Let A_i be the i-th student sits on seat which he has been sitting on Monday.

According to question,

We have to calculate P(\cap^{20}_{i=1}A_i^c)

Applying inclusion exclusion formula,

P(\cap^{20}_{i=1}A_i^c)=1-P(\cap^{4}_{i=1}A_i)

P(\cap^{20}_{i=1}A_i^c)=1-P(A_1)+...+P(A_{20})-P(A_1\cap A_2)+...+P(A_{19}\cap A_{20})+P(A_1\cap A_2\cap A_3)+...+P(A_{18}\cap A_{19}\cap A_{20})....-P(A_1\cap A_2...\cap A_{20})

Using symmetry,

P(\cap^{20}_{i=1}A_i^c)=1-\sum^{20}_{k=1}(-1)^{k+1}\binom{20}{k}P(A_1\cap ...\cap A_k)

P(\cap^{20}_{i=1}A_i^c)=1-\sum^{20}_{k=1}(-1)^{k+1}\binom{20}{k}\frac{(20-k)!}{20!}

P(\cap^{20}_{i=1}A_i^c)=1+\sum^{20}_{k=1}(-1)^{k}\frac{1}{k!}

P(\cap^{20}_{i=1}A_i^c)=\sum^{20}_{k=0}(-1)^{k}\frac{1}{k!}

P(\cap^{20}_{i=1}A_i^c)=\frac{1}{e}

Therefore, The probability that no one sits in the same seat on both days of that week is given by, P(\cap^{20}_{i=1}A_i^c)=\frac{1}{e}

8 0
3 years ago
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