Matt Wolf - CB East Wolf
... 1) Two coins are tossed. What is the probability that both land heads up? 2) A six-sided die is rolled. Calculate the following probabilities: a. P(rolling a 5) = b. P(rolling an even number) = c. P(rolling an odd number) = d. P(rolling a prime number) = 3) A card is chosen from a 52-card deck. Calc ...
... 1) Two coins are tossed. What is the probability that both land heads up? 2) A six-sided die is rolled. Calculate the following probabilities: a. P(rolling a 5) = b. P(rolling an even number) = c. P(rolling an odd number) = d. P(rolling a prime number) = 3) A card is chosen from a 52-card deck. Calc ...
Chapter 4
... Introduction • This chapter introduced the basic concepts of probability. It outlined rules and techniques for assigning probabilities to events. • One objective in his and the following chapters is to develop the probability-based tools that are at the basis of statistical inference. ...
... Introduction • This chapter introduced the basic concepts of probability. It outlined rules and techniques for assigning probabilities to events. • One objective in his and the following chapters is to develop the probability-based tools that are at the basis of statistical inference. ...
Basic things you need to know about sets and probability
... You may have noticed that when you complement a union or intersection, each set gets complemented and the operation switches (union becomes intersection and intersection becomes union). ...
... You may have noticed that when you complement a union or intersection, each set gets complemented and the operation switches (union becomes intersection and intersection becomes union). ...
Slides
... • If E and F are mutually exclusive events (E ∩ F = ∅) then P(E ∪ F ) = P(E ) + P(F ). Otherwise P(E ∪ F ) = P(E ) + P(F ) − P(E ∩ F ). • P(E c ) = 1 − P(E ) • E ⊂ F implies that P(E ) ≤ P(F ) • If all outcomes equally likely then P(E ) = ...
... • If E and F are mutually exclusive events (E ∩ F = ∅) then P(E ∪ F ) = P(E ) + P(F ). Otherwise P(E ∪ F ) = P(E ) + P(F ) − P(E ∩ F ). • P(E c ) = 1 − P(E ) • E ⊂ F implies that P(E ) ≤ P(F ) • If all outcomes equally likely then P(E ) = ...
15.6 Review Solutions
... 4) Write a verbal statement for the COMPLEMENT of each event. a. Drawing 2 cards from a standard deck. b. 6 question true or false test ...
... 4) Write a verbal statement for the COMPLEMENT of each event. a. Drawing 2 cards from a standard deck. b. 6 question true or false test ...
Probability theory and average
... – Rolling a 6-sided die and getting an even number (this is more than one outcome—3 to be exact!) – Drawing a card and getting an ace (4 outcomes!) ...
... – Rolling a 6-sided die and getting an even number (this is more than one outcome—3 to be exact!) – Drawing a card and getting an ace (4 outcomes!) ...
Fall 2002
... remember the honor code we discussed in class and show all of your work for each of the problems. You may use any printed materials, including other textbooks, but you must provide appropriate citations. Oral, written, electronic, or any other form of communication with other individuals is strictly ...
... remember the honor code we discussed in class and show all of your work for each of the problems. You may use any printed materials, including other textbooks, but you must provide appropriate citations. Oral, written, electronic, or any other form of communication with other individuals is strictly ...
Answers
... what is E[X(11 − X)]? What is V ar[X(11 − X)]? Solution: One might be tempted to approach the problem algebraically. That is, by expanding the expectation as follows: E(X(11 − X)) = E(11X) − E(X 2 ) = 11 × E(X) − E(X 2 ) and then computing E(X) and E(X 2 ) for the random variable in question. Howeve ...
... what is E[X(11 − X)]? What is V ar[X(11 − X)]? Solution: One might be tempted to approach the problem algebraically. That is, by expanding the expectation as follows: E(X(11 − X)) = E(11X) − E(X 2 ) = 11 × E(X) − E(X 2 ) and then computing E(X) and E(X 2 ) for the random variable in question. Howeve ...
