Say she chooses topic A. She needs at least one book to arrive on time. The probability that no books arrive on time is 0.1 x 0.1 = 0.01 so the probability that at least one book arrives on time is 1 - 0.01 = 0.99. Say she chooses topic B. She needs at least two books to arrive on time. The probability of no books arriving on time is 0.1 x 0.1 x 0.1 x 0.1 = 0.0001. The probability of exactly one book arriving on time is 4 x 0.1 x 0.1 x 0.1 x 0.9 = 0.0036 (binomial probability distribution formula - the 4 comes from 4C1). So the probability of her not getting enough books is 0.0001 + 0.0036 = 0.0037 and the probability she does get the books she needs on time is 1 - 0.0037 = 0.9963. So she should choose topic B.
We can use SOH, CAH and TOA, but first we must find the length of a hypotenuse: h = sin (theta)= 2/√13 cos ( theta) = -3/√13 tan (theta) = -2/3 cot (theta ) = -3/2 sec (theta) = 1/cos(theta ) = -√13/3 csc(theta ) = 1/ sin (theta) = √13/2
