document.write( "Question 697397: Two cards are selected, one at a time from a standard deck of 52 cards. Let x represent the number of fives drawn in a set of 2 cards.
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\n" ); document.write( "(A) If this experiment is completed without replacement, explain why x is not a binomial random variable.
\n" ); document.write( "(B) If this experiment is completed with replacement, explain why x is a binomial random variable \n" ); document.write( "
Algebra.Com's Answer #430016 by Positive_EV(69)\"\" \"About 
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For a variable to be a binomial random variable, it has to satisfy the requirement that all trials have the same probability of success. In this example, a \"success\" is a five being drawn.
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\n" ); document.write( "\n" ); document.write( "In case 1), The probability that the first card drawn is a five is the number of fives in the deck over the number of cards in the deck, or 4/52 = 1/13. If the five is not replaced, there are now 3 fives left in the deck, and 51 total cards. For the second draw, the probability of drawing a five is now 3/51 = 1/17. Since 1/13 is not equal to 1/17 the probabilities of success are not the same, and this is not a binomial distribution.
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\n" ); document.write( "\n" ); document.write( "Alternatively, if you did not draw a five, there are 4 fives left and 51 cards, so the probability the second card is a five is 4/51. Once again, the probability of the second card being a five is not the same as that of the first card being a five, so this is not a binomial distribution.
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\n" ); document.write( "\n" ); document.write( "In case 2), The probability of the first card being a five is still 4/52 = 1/13; however, since the five is replaced, for draw two there is once again 4 fives in a deck of 52 cards. Or, if you did not draw a five, that card is replaced, and there's 52 cards in the deck again, with 4 fives. Thus, the probability of drawing a second five is once again 4/52 = 1/13. Since there is a defined event which is considered a \"success\" (drawing a five), a defined number of trials (2), and the same probability of success for each trial (1/13), the numbers of fives drawn is a binomial random variable with n = 2 and p = 1/13. \n" ); document.write( "
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