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10 Feb 2021, 06:15
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63% (02:28) correct
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A students have to answer 10 out of 13 questions in an examination. The number of ways in which he can answer if he must answer at least 3 of the first five questions is
(A) 276
(B) 267
(C) 80
(D) 75
(D) 70
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
(A) 276
(B) 267
(C) 80
(D) 75
(D) 70
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
10 Feb 2021, 06:50
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5C3*8C7 + 5c4*8c6+ 5c5*8c5 = 276
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hykhan
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03 Apr 2021, 11:32
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So what we are trying to figure out is how many ways the student can answer 10 questions, if he must answer at least 3, and up to 5, out of the first 5 questions. An at least problem in probability should alert you that you need to add up the different distinct probabilities of each case.
First, figure out how many ways he can answer 10 out of 13 if he chooses 3 out of the first 5 questions, then 4 out of 5, then all 5 questions. Finally, add up the 3 different probabilities.
5C3 ways to choose 3 questions x 8C7 (which is 10 minus 3 answers and 13 minus 5 out of the remaining questions) = 10 x 8 = 80
+
5C4 ways to choose 4 questions x 8C6 (which is 10 minus 4 answers and 13 minus 5 out of the remaining questions) = 5 x 28 = 140
+
5c5 ways to choose 5 questions x 8C5 (which is 10 minus 5 answers and 13 minus 5 out of the remaining questions) = 1 x 56 = 56
80 + 140 + 56 = 276
First, figure out how many ways he can answer 10 out of 13 if he chooses 3 out of the first 5 questions, then 4 out of 5, then all 5 questions. Finally, add up the 3 different probabilities.
5C3 ways to choose 3 questions x 8C7 (which is 10 minus 3 answers and 13 minus 5 out of the remaining questions) = 10 x 8 = 80
+
5C4 ways to choose 4 questions x 8C6 (which is 10 minus 4 answers and 13 minus 5 out of the remaining questions) = 5 x 28 = 140
+
5c5 ways to choose 5 questions x 8C5 (which is 10 minus 5 answers and 13 minus 5 out of the remaining questions) = 1 x 56 = 56
80 + 140 + 56 = 276
