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
If this is a true or false question, it’s false because 2/10 + 7 ≠ 10. the correct inequality would be 2/10 + 7 = 12.
How do you say that having any confusion he said wouldn’t be able to consist of all the new stages
Step-by-step explanation:
2x+3=x+x+3
add the X's on the right side together.
2x+3=2x+3
subtract 2x from both sides
3=3
subtract 3 from both sides
0=0
the statement is true for any value of x
Use the Pythagoras Theorem in the corresponding right angled triangles.
