College Math

College Math Study Guide

The event of drawing candies and chocolates are independent of each other. The probability of drawing candies and chocolates respectively are given as: P(A) = 5/9 and P(B) = 5/9 The probability that both candies are drawn would be P(A)*P(B) ( ∩ B) = 5 9 ∗ 5 9 = 25 81 = 0.3086 It should be noted that we use union in case we have to find the probability of event A OR event B, while intersection is used when the probability of event A AND event B have to be found. Further, P(AUB) = P(A) + P(B) – P(A∩B) 4.10 Expected Value In real life situations, we have to find the expected value of the occurrence of an event. For instance, you play a game in which you roll a die. If you get a number less than 5, you lose, and if you get 5 or more, you win. In the case where you put money in the game and invest 5 dollars per play, you get nothing if you lose and get $20 if you win; if you play many times, you might be interested to know the expected value of winning. In this case, the probability of winning and losing has to be determined first. We know that a die has a total of 6 digits, and if we get any number between 1 and 4, we lose, and if we get 5 or 6, then we win. Hence, the winning probability is 2/6 and losing probability is 4/6. If we play a total of 300 games, then it is expected to win for about 100 times and lose in 200 times. The expected value of the total 300 games would be: 100*15 + (-5)*200 = 500 The expected value per game would be 500/300 = $1.67 Hence, it can be said that, when a game gives a payoff of a 1 , a 2 , …a n , with the respective probabilities of p 1 , p 2 ,…p n , then the expected value of the game is given by the following formula: = 1 1 + 2 2 + ⋯+ 4.11 Conditional Probability Conditional probability is the probability of an event, such that one event has already happened. For instance, two players are playing a game of cards, and the first player draws one card and the second

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