Questions and Answers #1 Probability

Questions and Answers Sheet 1 Probability Question #1 Find the probability P(E or F) if E and F are mutually exclusive, P(E)=0.41, and P(F)=0.47 . The probability P(E or F) is nothing. Answer: Step 1 Given information According to the given information, we have Event E and F are mutually exclusive, P(E) = 0.41 P(F) = 0.47 Step 2 Answer is given by, The probability P(E or F) is given by, We know that, If E and F are mutually exclusive, then P(E and F)=0 and, P(E or F)=P(E)+P(F)-P(E and F) P(E or F)=0.41+0.47-0 P(E or F)=0.88 Therefore, the probability P(E or F) is 0.88. Question #2 At a doctor's office, patients arrive at a rate of ten per hour. With a probability of 1 2 , each is either young or old. Assume you know that exactly ten young people arrived within an hour (say, 10 to 11am) a) Compute the probability that exactly 10 old patients also entered. b) Compute the probability that at least 20 patients have entered Answer: Step 1 Given information: Given that at a doctor’s office, patients arrive at a rate of 10 per hour with a probability of 0.5. Step 2 a) Compute the probability that exactly 10 old patients also entered: The probability that exactly 10 old patients also entered is obtained as given below: X= Number of old patients 𝑋 ∼ 𝑃(𝜆 = 0.5 × 10 = 5) 𝑃(𝑋 = 10) = = 65800.26366 𝑒 −5 ×510 10! 3628800 =0.0181 Thus, the probability that exactly 10 old patients also entered is 0.0181. Step 3 b) Compute the probability that at least 20 patients have entered: The probability that at least 20 patients have entered is obtained as given below: 𝑃(𝑋 ≥ 20) = 1 − 𝑃(𝑋 ≤ 19) = 1 − ∑19 𝑥=0 𝑒 −10 ×10𝑥 𝑥! =1-0.9965 =0.0035 Thus, the probability that at least 20 patients have entered is 0.0035. Question #3 Which number has an unlikely probability? 7 A) 8 B) 0 1 C) 4 D) 1 Answer: Step 1 1 An unlikely event would have a probability between 0 and . 2 1 A likely event would have a probability between and 1. 2 Step 2 Here 0 and 1 4 are unlikely probability. Therefore, Option B and C are correct. Question #4 If the probability of winning a game of chance is 2 13 , what is the probability of not winning the game? Answer: Step 1 Probability: Probability is a field of mathematics that gives us the language and tools to quantify the uncertainty of events and reason in a principled manner. It is a way of measuring the chance of something happening. An event is any occurrence that has a probability attached to it. In other words, an event is any outcome where you can say how likely it is to occur. Probability is measured on a scale of 0 to 1. If an event is impossible, it has a probability of 0. If it’s an absolute certainty, then the probability is 1. Step 2 Define events: A : Winning a game B : Losing a game (not winning a game) 2 Given 𝑃(𝐴) = 13 Total probability =P(A)+P(B) 2 ⇒ 1 = + 𝑃(𝐵) 13 ⇒ 𝑃(𝐵) = 1 − ∴ 𝑃(𝐵) = 10 2 13 = 10 13 13 Probability of not winning the game = 10 13 . Question #5 Answer the following conceptual questions about probability. a) Suppose your friend Rudy says that the probability it rains tomorrow is 125 percent. Is Rudy’s statement coherent? Explain your answer. b) Suppose Rudy assigns probability 0.3 to the sentence (𝑃 ⇒ 𝑄). What probability should Rudy assign to the sentence (∼ 𝑃 ∨ 𝑄)? Explain your answer. c) Suppose Rudy buys a lottery ticket. You say, “Don’t you know how unlikely it is that you’ll win?” Rudy replies, “Either I’ll win or I won’t, so there’s a 50% chance!” Is Rudy correct? What might you say to Rudy in this case? Answer: Step 1 a) This statement is incorrect as the probability can never be greater than 1 or 100%. Step 2 b) If probability of (𝑃 ⇒ 𝑄) = 0.3, then, Probability of (∼ 𝑃 ∨ 𝑄) = 1 − 0.3 = 0.7 Step 3 c) The statement would be true if there would have been only two persons purchasing the lottery out of which one will win. Since there are so many people purchasing the lottery ticket and only a few will win, hence the probability is very small. Question #6 a) What is 𝑃(3 < 𝑋 ≤ 10) ? b) What is 𝑃(𝑋 ≥ 3) ? Given the discrete probability distribution of a certain experiment, answer the following questions. x 1 3 4 5 10 Probability P ( X = x ) 0.30 0.20 0.10 0.15 0.25 Answer: Given data, - The probability distribution table of the certain experiment is, x 1 3 4 5 10 Probability P ( X = x ) 0.30 0.20 0.10 0.15 0.25 Step 1 a) The value of 𝑃(3 < 𝑋 ≤ 10) is, 𝑃(3 < 𝑋 ≤ 10) = 𝑃(𝑋 = 4) + 𝑃(𝑋 = 5) + 𝑃(𝑋 = 10) The probability of X=4 is 0.10. The probability of X=5 is 0.15. The probability of X=10 is 0.25. Substituting the values then, 𝑃(3 < 𝑋 ≤ 10) = 𝑃(𝑋 = 4) + 𝑃(𝑋 = 5) + 𝑃(𝑋 = 10) =0.10+0.15+0.25 =0.50 Hence, the required probability is 0.50. Step 2 b) The value of 𝑃(𝑋 ≥ 3) is, 𝑃(𝑋 ≥ 3) = 𝑃(𝑋 = 1) + 𝑃(𝑋 = 3) The probability of X=1 is 0.30. The probability of X=3 is 0.20. Substituting the values then, 𝑃(𝑋 ≥ 3) = 𝑃(𝑋 = 1) + 𝑃(𝑋 = 3) =0.30+0.20 =0.50 Hence, the required probability is 0.50. Question #7 Suppose that the random variable x, shown below, represents the number times. P(x) represents the probability of a randomly selected person having received that number of speeding tickets during that period. Use the probability distribution table shown below to answer the following questions. x 0 1 2 3 4 5 6+ P( x) 0.3176 0.2705 0.2469 0.0832 0.0485 0.0333 0.0000 a) What is the probability that a randomly selected person has received four tickets in a three-year period? P(x=4)=P(x=4)= b) What is the probability that a randomly selected person has received three tickets in a three-year period? P(x=3)=P(x=3)= c) What is the probability that that a randomly selected person has received more than one tickets in a three-year period? P(x>1)=P(x>1)= d) What is the probability that that a randomly selected person has received two or less tickets in a threeyear period? 𝑃(𝑥 ≤ 2) = 𝑃(𝑥 ≤ 2) = Answer: Step 1 Part A The probability that a randomly selected person has received four tickets in a three-year period is given from probability distribution table as: P(X=4)=0.0485 Part B The probability that a randomly selected person has received three tickets in a three-year period is given as: P(X=3)=0.0832 Step 2 Part C The probability that that a randomly selected person has received more than one tickets in a three-year period is given as: P(X>1)=1-(P(X=0)+P(X=1)) =1-(0.3176+0.2705) =1-0.5881 =0.4119 Part D The probability that a randomly selected person has received two or less tickets in a three-year period is given as: 𝑃(𝑋 ≤ 2) = (𝑃(𝑋 = 0) + 𝑃(𝑋 = 1) + 𝑃(𝑋 = 2)) =0.3176+0.2705+0.2469 =0.835 Question #8 if the probability of event A is 3 4 and the probability of event B is 1/4 then the probability of both A and B occuring at the same time is A)1 or absolute certainty 1 B) C) D) E) 2 3 16 3 4 1 4 Answer: Step 1 Given information Probability of event A is Probability of event B is 3 4 1 4 Step 2 Answer is given by, The probability of both A and B occurring at the same time is given by, P(A and B)=P(A)P(B) 3 1 3 P(A and B) = × = 4 4 16 The probability of both A and B occurring at the same time is 3 16 Option (C) is correct. Question #9 A standard deck of cards contains 52 cards. One card is selected from the deck. a) Compute the probability of randomly selecting a spade or diamond. b) Compute the probability of randomly selecting a spade or diamond or heart. c) Compute the probability of randomly selecting a king or club. a. 𝑃(spade or diamond) =? b. 𝑃(spade or diamond or heart) =? c. 𝑃(king or club) =? Answer: Step 1 a. In a standard deck of playing cards there are total 52 cards and 4 suits, such as hearts, diamonds, spades and clubs. Each of the suit contains 13 cards. Step 2 13 The probability of selecting a spade is 52 The probability of selecting a diamond is 1 = . 4 13 1 = . 52 4 The selection of spade and diamond are independent of each other. Hence, the probability of randomly selecting a spade or a diamond is, 1 4 1 1 4 2 + = = 0.500. Step 3 b. The probability of selecting a heart is 13 52 1 = . 4 The selection of spade or diamond or heart are independent of each other. 1 1 1 3 Hence, the probability of randomly selecting a spade or a diamond is, + + = = 0.750. 4 4 4 4 Step 4 c. The probability of selecting a club is 13 52 1 = . 4 Each if the suit contains a king. Hence, there are total 4 king in a deck of cards. 4 1 The probability of selecting a king is = . 52 13 There is only one card, which is king of club. Thus, the probability of selecting a king of club is Thus, the probability of randomly selecting a king or club is, 1 4 + 1 13 − 1 52 = 4 13 = 0.308. Question #9 Let X be a random variable with range [0,1] and probability density function (pdf) is given as: 1 𝑓(𝑋) = 3𝑥 2 . Find probability 𝑃 (𝑋 > ). 5 Answer: Step 1 Let X be a random variable with range [0,1]and probability density function (pdf) is given as: 𝑓(𝑋) = 3𝑥 2 1 We need to find the probability 𝑃 (𝑋 > ) 5 Here we can write probability density as, 3 x 2 , 0  x  1 f (X ) =  otherwise 0 Step 2 1 1 𝑃 (𝑋 > ) = ∫1/5 𝑓(𝑥)𝑑𝑥 5 1 ∫1/5 3𝑥 2 𝑑𝑥 1 = 3 ∫1/5 𝑥 2 𝑑𝑥 𝑥3 1 = 3[ ] 3 1/5 (1/5)3 1 = 3[ − 3 1 1 3 375 = 3[ − 3 ] ] =1-0.008 =0.992 Hence, the probability 𝑃(𝑋 > 1/5) is 0.992. 1 52 .