Expected value (Also Mean): mu_x = sigma (x middot P(x)) Standard Deviation: sig
ID: 3226641 • Letter: E
Question
Expected value (Also Mean): mu_x = sigma (x middot P(x)) Standard Deviation: sigma_x = Squareroot sigma (x^2 middot P(x)) - mu_x^2 (CAUTION: mu_x^2 is NOT part of the summation). Binomial Probability Distribution Formula: P(x) = _nC_x(p)^x(1 - p)^n - x Number possible combinations of selecting x items given a total of n items: _nC_x = n!/x!(n - x)! Expected Value = sigma (P(x) middot A(x)), where each x is a possible outcome, A(x) is the numerical result of each outcome (could be positive, negative, or 0), P(x) is the probability of each outcome, and of course sigma P(x) = 1. There are 52 cards in a standard deck of playing cards, 4 clubs, 4 diamonds, 4 hearts, and 4 spades. There are 13 aces, 13 kings, 13 queens, 13 jacks, 13 tens, etc. Two cards a selected at random from a deck. a) What is the probability of picking 2 aces, without replacement? b) What is the probability of picking an ace and then a king, without replacement? c) What is the probability of picking 2 aces, with replacement? d) What is the probability of picking and ace and then a king, with replacement? e) What is the probability of picking one card that is both an ace and a diamond? f) What is the probability of picking one card that is both an ace and a king?Explanation / Answer
a) Probability of picking 2 aces without replacement
n(s)=52
A= event of drawing first aces P(A)=13/52 without replacement we have 12 aces from 51 cards
B= event of drawing second aces P(BIA)=12/51
hence Probability of picking 2 aces without replacement= P(A)*P(BIA) = (13/52)*(12/51) =0.0588
b) Probability of picking 1 ace then 1 king without replacement
n(s)=52
A= event of drawing first ace P(A)=13/52 without replacement we have 51 cards left
B= event of drawing second card king P(BIA)=4/51 if first selected card is king of ace then P(BIA)'=3/51
hence Probability of picking 1 ace then 1 king without replacement= P(A)*(P(BIA) +P(BIA)')= (13/52)*((4/51)+(3/51)) = 0.0343
a) Probability of picking 2 aces with replacement
n(s)=52
A= event of drawing first aces P(A)=13/52 with replacement
B= event of drawing second aces P(B)=13/52
hence Probability of picking 2 aces with replacement= P(A)*P(B) = (13/52)*(13/52) =0.0625
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.