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.
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
The fancy or strange looking E symbol is the greek uppercase letter sigma, to mean sum. So they want you to add the terms in the form x^2. The x values are from the list given to us.
x = 16 leads to x^2 = 256
x = 13 leads to x^2 = 169
etc etc
After squaring all the x values, we have this new list {256, 169, 25, 400, 196}
Lastly, you add up all the squared values to get the final answer of 256+169+25+400+196 = 1046
