Out of 17 applicants 8 boys and 9 girls. Two persons are to be selected for the job. Find the probability that at least one of the selected persons will be a girl.
Answer: B The events of selection of two person is redefined as first is a girl and second is a boy or first is boy and second is a girl or first is a girl and second is a girl. i.e (8/17 * 9/16) + (9/17 * 8/16) + (8/17 * 7/16) = 9/34 + 9/34 + 7/34 = 25/34.
Q. No. 20:
Two cards are drawn from a pack of well shuffled cards. Find the probability that one is a club and other in King.
Answer: D Let X be the event that cards are in a club which is not king and other is the king of club. Let Y be the event that one is any club card and other is a non-club king. Hence, required probability = P(A) + P(B) = (12C1*1C1)/52C2 + (13C1*3C1) /52C2 =>{2(12+1)/(52*51)} + {2(13*3)/(52*51)} = (24+78)/(52*51) = 1/26
Q. No. 21:
Two teams Arrogant and Overconfident are participating in a cricket tournament. The odds that team Arrogant will be champion is 5 to 3, and the odds that team Overconfident will be the champion is 1 to 4. What are the odds that either Arrogant or team Overconfident will become the champion?
Answer: D As probability of a both the teams (Arrogant and Overconfident) winning simultaneously is zero. P(A ∩ O) = 0 P(A ∪B) = P(A) + P(B) = 5/8 + 1/5 = 33/40 So required odds will be 33 : 7
Q. No. 22:
A five-digit number is formed by using digits 1,2,3,4 and 5 without repetition. What is the probability that the number is divisible by 4 ?
Answer: A A number divisible by 4 formed using the digits 1,2,3,4 and 5 has to have the last two digits 12 or 24 or 32 or 52. In each of these cases, the five digits number can be formed using the remaining 3 digits in 3! = 6 ways. A number divisible by 4 can be formed in 6*4 = 24 ways. Total number that can be formed using the digits 1,2,3,4 and 5 without repetition => 5! = 120 Required probability = 24/120 = 1/5.
Q. No. 23:
A special lottery is to be held to select a student who will live in the only deluxe room in a hostel. There are 100 Year-III, 150 Year-II and 200 Year-I students who applied. Each Year-III's name is placed in the lottery 3 times; each Year-II's name, 2 times and Year-I's name, 1 time. What is the probability that a Year-III's name will be chosen ?
Answer: A The probability that exactly 4 vessels arrive safely is : 5c4*(9/10)4 (1/10) The probability that all 5 arrive safely is (9/10)5 The probability that atleast 4 vessels arrive safely = 5c4*(9/10)4 (1/10) + (9/10)5 = (14*94 )/105