Question

In a high school graduating class of 100 students, 54 studied mathematics, 69 studied history, and 35 studied both mathematics and history. If one of these students is selected at random, find the probability that (a) the student took mathematics or history; (b) the student did not take either of these subjects; (c) the student took history but not mathematics. . . .

Answer 1

1) Given, In a class of 100 students , s4 study mathematics (M), 69 study History (H), and 35 study both mathematics and History n (m)=54 (H)=69 n (mnH)=35 P(M)= sy a 0.54 P(H) = 0.69 (mnha 35 = 0.35 a) Probability that the student took mathemales or P(mUH) = P(M) + P(HD - P(mnh) - 0.54 + 0.69 -0.35 = 0.88 History: b) The probability that the student did not lake either of There classes: P (mind) = 1- P(mUH) = 1- 0.88 = 0.12

student took history but The probability that a The not mathemalics: P(min H) = P(H)- P (MNH) - 0.8 0.35 = 0.34 t n be the event " that that there a are a married m married man walches the show that a married woman walches let The in be the event show The probability that a married man watches the show is o.4 - P(m) = 0.4 that a married woman watches the show, The is probability 0.5 - P(w)= o.s. The probabilila lhat a man watches the he show.. show , given that his wife does, is 0.7 P(m/w) = 0.7

" The probability that A married couple walches the show P(m nw)= = P(W) P(m/w) (0.5) (0.7) - 0.35 3 The show, given. The probability That a wife walches that her husband does is: P(w lm) = P(wom) P(m) 0.35 = 0.875 0.4 c) The probability that at least one member of o married couple will watch the show: P(wum) = P(W)+ s(m) - P(wnm) = 0.5+ 0.4-0.35 - 0.55 please rate if it helps.