Probability Question & Answers

Probability Question & Answers (Goal Sept Excel) In a recent street fair students were challenged to hit one of the shaded triangular regions on the large equilateral triangular board below with a ping pong ball. Each of the triangular regions is an equilateral triangle whose side is a third of the length of the large triangle board. If the ping pong ball hits the large triangular region, what is the probability of hitting a shaded triangle? 1/5 1/4 1/3 1/2 2/3 - ANSWER: C Correct. This is a geometric probability question. Remember the formula for geometric probability: In this case, the issue is area of the three black triangles. You are asked to find the area of the small black triangles out of the area of the big triangle (the entire shape). In order to avoid unnecessary calculations, try to divide the hexagon into equilateral triangles - you only need the ratio between the shaded triangles and the unshaded ones. Remember - GMAT figures are drawn to scale unless otherwise stated. Divide the unshaded hexagon into 6 identical equilateral triangles, by drawing the main diagonals:The black triangles are identical to the 6 triangles of the hexagon - their angles must also be 60º (in order to supplement the hexagon's angles, 120º, and they each share a side with the hexagon's triangle. They even look the same area.) Thus, the entire shape is made of 9 identical triangles, 3 of them shaded. Therefore, the chance of hitting them in a random throw of a ball is 3/9 = 1/3. Avoid unnecessary calculations - ballpark! 2 people are to be selected from Abraham, Benjamin, Chris, and Dave. What is the probability that both Abraham and Benjamin will be selected? 1/12 1/6 1/3 2/3 5/6 - ANSWER: A Incorrect. This question presents two possible scenarios to reach the desired outcome. Identify these scenarios, calculate each one using the following method, then add the probabilities of all scenarios, since they are connected by an "OR" relationship. Within each scenario: Break down the pair of chosen persons into separate, single events (first choice, second choice); calculate the probability for each event One-at-a-time; then MULTIPLY the probabilities, as the events have an AND relationship between them. When considering each event one-at-a-time, assume that the previous events yielded a wanted response, and consider each event under the new conditions created by this assumption. B Correct. The two wanted scenarios are: Abraham is chosen first - 1/4, and then Benjamin is chosen - 1/3. a total probability of 1/4×1/3 = 1/12 OR Benjamin is chosen first - 1/4, and then Abraham is chosen - 1/3. a total probability of 1/4×1/3 = 1/12 Since this is an "OR" relationship, add the probabilities: 1/12+1/12 = 2/12 = 1/6 Alternative method using combinations: When calculating complex probabilities, start from the denominator. It is the number of all pairs. This is equal to picking 2 out of 4, not ordered, with no repetition, since you cannot pick the same person twice. C(4,2) = 6 Abraham and Benjamin are 1 pair, so the probability that they will be chosen is 1/6 A bag contains 3 red, 4 black and 2 white balls. What is the probability of drawing a red ball and a white ball (not necessarily in that order) in two successive draws, each ball being put back after it is drawn? 2/27 1/9 1/3 4/27 2/9 - ANSWER: A Incorrect. This is a multiple event probability question - two events of drawing a ball. Break down the question into separate, single events; calculate the probability for each event One-at-a-time; then MULTIPLY the probabilities, as the events have an AND relationship between them. When considering each event one-at-a-time, assume that the previous events yielded a wanted response, and consider each event under the new conditions created by this assumption. When the balls are put back, each event starts with all items, and the conditions do not change. Note: This is the probability of drawing a red ball first and a white ball second, but what about drawing a white ball first and a red ball second? D Correct. There are 2 good scenarios in this question: 1) Drawing a red ball, putting it back and drawing a white ball: There are 3 red balls and 3+4+2 = 9 balls in total, thus the probability of drawing a red ball is 3/9 = 1/3. Now put back the red ball. There are 2 white balls out of 9, thus the probability of drawing a white ball is 2/9. Total probability = (1/3)×(2/9) = 2/27 OR 2) Drawing a white ball, putting it back and drawing a red ball: There are 2 white balls and 3+4+2 = 9 balls in total, thus the probability of drawing a red ball is 2/9 Now put back the white ball. There are 3 red balls out of 9, thus the probability of drawing a red ball is 3/9 = 1/3. Total probability = (1/3)×(2/9) = 2/27. Note that the order of drawing the balls does not change the probability. Since There's an OR relationship between scenarios, add: 2/27 + 2/27 = 4/27 A basic model of a slot machine randomly shuffles and presents different combinations of the letter of the word MATH. What is the probability that the position of the "A" remains unchanged when the letters of the word MATH are re-arranged? 1/4 1/6 1/3 1/24 1/12 - ANSWER: C............................