Two cards are drawn at random from a single deck of playing cards. (a) What is t

Question

Two cards are drawn at random from a single deck of playing cards. (a) What is the probability of drawing the queen of diamonds followed by the king of diamonds? (b) What is the probability of drawing a queen and an ace in either order? A computer program randomly generates a codeword consisting of two letters (from a - z) followed by 2 digits from (0-9). Letters can be repeated in the same codeword, but a number can appear only once in a codeword. (a) What is the probability that the code word has only one vowel (a, e, i, o, u)? (b) What is the probability that the code word has two vowels? (c) What is the probability that the code word ends with the last two digits of your Student ID, if all numbers are equally likely to be generated by the program. (d) What is the probability that the code word ends with the last two digits of your Student ID, if odds are three times as likely to be generated by the program as evens. An architectural firm has two senior architects. The First Architect leads 70% of the company's projects. The Second Architect leads the other 30% of the company's projects. The board of directors meet each fiscal year and see which projects went over budget, and thus are considered losses for the firm. From experience, the company knows that 2% of the First Architect's projects go over budget while, for the Second Architect, 4% of his projects goes over budget. Suppose the board of directors found a project that went over budget. Which architect would you say did the work? Justify your answer. Three binary bits are combined together to form a symbol. If the probability of a "0" is equal to the probability of a "1", what is the probability that (a) the symbol will be either all zeros or all ones? (b) the symbol will have at least one "0" and atleast o "1"?

Explanation / Answer

1:

(a)

Out of 52 cards, one card is queen of diamonds and one is king of diamond.

So the probability of selecting queen of diamond first is

P(queen of diamond) = 1/52

After selecting this number of cards remaining is 51 so

P(king of diamond| queen of diamond) = 1/51

Therefore

P(king of diamond and queen of diamond ) = P(king of diamond| queen of diamond)*P(queen of diamond) = (1/51)*(1/52) = 0.000377

(b)

There are 4 queens and 4 aces so number of ways of selecting one queen and one ace is C(4,1)C(4,1) = 16

And number of ways of selecting 2 cards out of 52 is C(52,2) = 1326

So required probability is 16 /1326 = 0.0121

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