Assuming the two students are independent, meaning they did not work together in anyway shape or form when taking the course then both students have a 0.40 probability of failing the course. the probability that both students had failed is the product of 0.40 * 0.40 = 0.16

if the students worked together, then they are not independent and this is not the correct solution. Use this equation instead, which always works, P(A and B) = P(A) + P(B) – P(A or B)

From the tone of your question curiously that we are to think of that the risk of ingesting alcohol is an independant variable from the probablity of smoking weed. it fairly is probable no longer the case nonetheless. yet i can tell because of the fact the probablity given for doing the two is what you may assume if the two have been independant and had no correlation. this is, the probablity of no longer smoking weed on my own is .seventy six and the probablity of no longer ingesting is .40 4. So in the event that they’re independant then the risk of no longer doing the two may be the RSS (Root Sum Squared) of the two that’s .89. So the risk of doing the two is .12 that’s on the fringe of your .11 given. So with those being independant (back, it fairly is probable a incorrect assumption, yet is indicated via the question if we’ve a a million% margin of blunders) then permit’s locate the risk of doing neither. Now 40 4% does not drink. Of that group seventy six% does not smoke. So we’ve 33% that does neither. something do the two or the two (smoke OR drink) that’s sixty seven%. .sixty seven is the respond.

Going back 20 years to school maths, I’d say a 16% chance;

40% x 40% = 16%

there should be a 48% chance of one fail and one pass

(40%x60%=24%)+(60%x40%=24%)= 48%

and a 36% chance of both having passed

60%x60%=36%.