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question:Jeff decides to play with a Magic 8 Ball. Each time he asks it a question, it has a 2/5 chance of giving him a positive answer. If he asks it 5 questions, what is the probability that it gives him exactly 2 positive answers?
answer:There are binom{5}{2}=10 ways to choose 2 of 5 questions to give a positive response to. For each choice, there is a probability of left( frac{2}{5} right)^2 left( frac{3}{5} right)^3 that those 2 questions will yield a positive answer and the other 3 will not. Therefore, the total probability that exactly 2 of questions give a positve answer is 10 left( frac{2}{5} right)^2 left( frac{3}{5} right)^3 = boxed{frac{216}{625}}.
question:Alice and Bob play a game with a baseball. On each turn, if Alice has the ball, there is a 1/2 chance that she will toss it to Bob and a 1/2 chance that she will keep the ball. If Bob has the ball, there is a 2/5 chance that he will toss it to Alice, and if he doesn't toss it to Alice, he keeps it. Alice starts with the ball. What is the probability that Alice has the ball again after two turns?
answer:We can solve this problem by dividing it into cases. If Alice tosses the baseball to Bob on the first turn, there is a 2/5 chance that Bob will toss it back to her on the next turn. On the other hand, if Alice keeps the baseball on the first turn, there is a 1/2 chance that she will also keep it on the second turn. The total probability is then frac{1}{2}cdotfrac{2}{5} + frac{1}{2}cdotfrac{1}{2}=boxed{frac{9}{20}}.
question:Given that binom{17}{7}=19448, binom{17}{8}=24310 and binom{17}{9}=24310, calculate binom{19}{9}.
answer:We can apply Pascal's identity to get that binom{19}{9}=binom{18}{8}+binom{18}{9}. From here, we can apply it twice more to get that binom{19}{9}=binom{18}{8}+binom{18}{9}=left(binom{17}{7}+binom{17}{8}right)+left(binom{17}{8}+binom{17}{9}right). Substituting the provided values of binom{17}{7}, binom{17}{8}, and binom{17}{9} gives us binom{19}{9}=19448+2(24310)+24310=boxed{92378}.
question:w many different routes are there from point A to point B if you can only move to the right or down along the drawn segments? [asy] unitsize(0.09inch); draw((0,0)--(10,0)--(10,10)--(0,10)--cycle); draw((5,0)--(5,10)); draw((0,5)--(10,5)); dot((0,10)); dot((10,0)); label("A",(0,10),NW); label("B",(10,0),SE); [/asy]
answer:To get from A to B, four moves are required: two down and two to the right. This problem can be thought of as asking how many ways there are to arrange the order of the four moves. Think of a downward move as the letter ``D" and a move to the right as the letter ``R". So, we are trying to count the total number of four-letter words formed with two Ds and two Rs. To do so, we can simply count the number of ways to arrange the Ds (the other slots will be automatically filled in with Rs). So, there are 4 slots in which the first D could go, and three in which the second could. However, we must divide by 2 for over-counting because the Ds are indistinct. Thus, there are frac{4 cdot 3}{2} = boxed{6} different routes.