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Burka [1]
4 years ago
6

ramone has 5 difficult questions left to answer on a multiple choice test. Each question has 3 choices. For the first 2 of these

questions, he eliminated 1 of the 3 choices. find the probability that he will answer the first 2 questions, as well as at least 2 of the 3 remaining questions correctly.
Mathematics
1 answer:
olchik [2.2K]4 years ago
5 0
1. a b c
2. a b c
3. a b c
4. a b c
5. a b c

then she eliminated 1 choice in 1 and 2, say as follows

1.    b c
2. a b 
3. a b c
4. a b c
5. a b c

Probability of answering correctly the first 2, and at least 2 or the remaining 3 is 
P(answering 1,2 and exactly 2 of 3.4.or 5.)+P(answering 1,2 and also 3,4,5 )

P(answering 1,2 and exactly 2 of 3.4.or 5.)=
P(1,2,3,4 correct, 5 wrong)+P(1,2,3,5 correct, 4 wrong)+P(1,2,4,5 correct, 3 wrong)
also P(1,2,3,4 c, 5w)=P(1,2,3,5 c 4w)=P(1,2,4,5 c 3w )
so  
P(answering 1,2 and exactly 2 of 3.4.or 5.)=3*P(1,2,3,4)=3*1/2*1/2*1/3*1/3*2/3=1/4*2/9=2/36=1/18

note: P(1 correct)=1/2
         P(2 correct)=1/2
         P(3 correct)=1/3
         P(4 correct)=1/3
         P(5 wrong) = 2/3

P(answering 1,2 and also 3,4,5 )=1/2*1/2*1/3*1/3*1/3=1/108

Ans: P= 1/18+1/108=(6+1)/108=7/108
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Answer:

There are 70 sheep in the farm.

Step-by-step explanation:

Consider the provided information.

Let c represents the cows, s represents the sheep and p represents the pigs.

The number of cows and the number of sheep are in the ratio 6:5.

\dfrac{c}{s} =\dfrac{6}{5}

c =\dfrac{6s}{5}

The number of sheep and the number of pigs are in the ratio 2:1.

\dfrac{s}{p} =\dfrac{2}{1}

p =\dfrac{s}{2}

The total number is 189.

c+s+p=189

Substitute the value of c and p in above.

\dfrac{6s}{5}+s+\dfrac{s}{2}=189

\dfrac{12s+10s+5s}{10}=189

27s=1890

s=70

Hence, there are 70 sheep in the farm.

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Step-by-step explanation:

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Hope I helped yo.

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