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
To add monomials, you have to look at the variables that are accompanied by their coefficients. In the given problem, (–4c2 + 7cd + 8d) + (–3d + 8c2 + 4cd), you can combine both cd ut nt cd and c² and cd and d and d and c² because they have different variables.
<span>(–4c2 + 7cd + 8d) + (–3d + 8c2 + 4cd)
(-4c</span>² + 8c²) + (7cd + 4cd) + (8d - 3d)
4c² + 11cd + 5d
