WBJEE 2014 Solved Mathematics Question Paper – Part 15
Find WBJEE 2014 Solved Mathematics Question Paper – Part 15 in this article. This paper consists of 5 questions (#71 to #75 ) from WBJEE 2014 Mathematics paper. Detailed solution of these questions has been provided so that students can match their solutions.
About WBJEE Exam
WBJEE is a common entrance examinations held at state level for admission to the undergraduate level engineering and medical courses in the state of West Bengal. The Mathematics section of WBJEE 2014 engineering entrance exam consists of 80 questions.
Importance of Previous Years’ Paper:
Previous years’ question papers help aspirants in understanding exam pattern, question format, important topics and assessing preparation. It has also been seen that sometimes questions are repeated in WBJEE Exam. So, this paper will certainly boost your confidence.
(A) is equal to zero
(B) lies between 0 and 3
(C) is a negative number
(D) lies between 3 and 6
Clearly it is a negative number.
72. A student answers a multiple choice question with 5 alternatives, of which exactly one is correct. The probability that he knows the correct answer is p, 0 < p < 1. If he does not know the correct answer, he randomly ticks one answer. Given that he has answered the question correctly, the probability that he did not tick the answer randomly, is
Let us suppose that,
The probability that the student knows the answer = Y
The probability that the student do not know the answer = N
The probability that student choose correct option = C
73. A poker hand consists of 5 cards drawn at random from a well-shuffled pack of 52 cards. Then the probability that a poker hand consists of a pair and a triple of equal face values (for example 2 sevens and 3 kings or 2 aces and 3 queens, etc.) is
Total number of combination of 5 cards out of 52 cards = C(52, 5)
Consider 4 cards of the same face value,
Total number of combination of 2 cards out of 4 cards = C(4, 2)
Total number of combination of 2 cards out of 4 cards = C(4, 3)
Considering two sets of 4 cards with equal face values, the possible combination is
C(4, 2). C(4, 3)
There are 13 sets of so, total combination = C(13, 2)⋅C(4, 2)⋅C(4, 3)
Now, for each pair of ranks, we can have a pair of cards from the first rank and a triple from the second rank, or a triple from the first rank, and a pair from the second rank. So, we should multiply by 2.
So, the probability that a poker hand consist of a pair and a triple of equal face values is
Ans : (C)