What is the probability that a card selected from a deck will be either an ace or a spade?
A. 2/52
B. 2/13
C. 7/26
D. 4/13
E. 17/52
(13+3) / 52 = 4/13
Answer (D)
What is the probability that a card selected from a deck will be either an ace or a spade?
A. 2/52
B. 2/13
C. 7/26
D. 4/13
E. 17/52
(13+3) / 52 = 4/13
Answer (D)
Good... thats correct...
Explain your approach so that everyone can understand.
One More Probability sum.
6 persons seat themselves at round table. What is the probability that 2 given persons are adjacent?
(A) 1/5
(B) 2/5
(C) 1/10
(D) 1/7
(E) 2/15
What is the probability that a card selected from a deck will be either an ace or a spade?
A. 2/52
B. 2/13
C. 7/26
D. 4/13
E. 17/52
p(a u s) = p(a) + p(s) - p(a n s)
So,
4C1/52C1 + 13C1/52C1 - 1/52 = 16/52 = 4/13
What is the probability that a card selected from a deck will be either an ace or a spade?
A. 2/52
B. 2/13
C. 7/26
D. 4/13
E. 17/52
hi,
the number of ways to choose a card from a pack of 52 = 52
the number of ways a spade can be chosen = 13
the number of ways an ace can be chosen = 4
but one case counts an ace which is a spade too...
hence the required number of ways to choose either an ace or a spade = 13+4-1 = 16
required probability = 16/52 = 4/13
One More Probability sum.
6 persons seat themselves at round table. What is the probability that 2 given persons are adjacent?
(A) 1/5
(B) 2/5
(C) 1/10
(D) 1/7
(E) 2/15
Number of ways in which 6 people can sit at a round table is: 5!
Two of them to stick together, they need to considered as 1 unit, with inter-changeability, so 4! * 2!
Overall probability therefore is: 4! * 2! / 5! = 2/5
One More Probability sum.
6 persons seat themselves at round table. What is the probability that 2 given persons are adjacent?
(A) 1/5
(B) 2/5
(C) 1/10
(D) 1/7
(E) 2/15
hi,
the number of ways to arrange 6 persons on 6 seats on a round table = (6-1)! = 5!
the number of ways that the 2 persons are adjacent
(considering the 2 persons as a single entity and they can permute between themselves also) =
(5-1)!*2 = 4!*2
the required probability = 4!*2 / 5! = 2/5
hope i am correct...
Number of ways in which 6 people can sit at a round table is: 5!
Two of them to stick together, they need to considered as 1 unit, with inter-changeability, so 4! * 2!
Overall probability therefore is: 4! * 2! / 5! = 2/5
You too got it right
hi,
the number of ways to arrange 6 persons on 6 seats on a round table = (6-1)! = 5!
the number of ways that the 2 persons are adjacent
(considering the 2 persons as a single entity and they can permute between themselves also) =
(5-1)!*2 = 4!*2
the required probability = 4!*2 / 5! = 2/5
hope i am correct...
Yes you are...
While we are at Probability, Try this one:
Nine people including three couples are to be seated in a row of nine chairs. What is the probability that none of the couples are sitting together ?
A) 1/63
B) 65/126
C) 13/63
D) 61/126
E) 6/21
spade can be chosen in = 13 ways (this includes 1 ace too)
an ace can be chosen = 3 ways (one is already done under spades)
total cards - 52
so, (13 + 3) / 52 => 4/13
Good... thats correct...
Explain your approach so that everyone can understand.
While we are at Probability, Try this one:
Nine people including three couples are to be seated in a row of nine chairs. What is the probability that none of the couples are sitting together ?
A) 1/63
B) 65/126
C) 13/63
D) 61/126
E) 6/21
hi vikram,
this seemed to be an easy one to me but i got stuck in it...
and finally i has to ask for help from the cat quant thread (as it is quite hyper active
)so here's the solution to the question...
http://www.pagalguy.com/forum/quantitative-questions-and-answers/42674-official-quant-thread-cat-09-a-554.html#post1641348
and here's the concept behind it...
http://www.pagalguy.com/forum/quantitative-questions-and-answers/42674-official-quant-thread-cat-09-a-554.html#post1641365
PS: i apologize for my be-imani here...:)
Puys .. plz clarify..Only integers can be even or odd??
For eg: 6^1/2 , 8^1/2( wat bout des no???)
( ^- sqr root)
thanks
For any integer k > 1, the term length of an integer refers to the number of positive prime factors, not necessarily distinct, whose product is equal to k. For example, if k = 24, the length of k is equal to 4, since 24 = 2 2 2 3. If x and y are positive integers such that x > 1, y > 1, and x + 3y
A. 5
B. 6
C. 15
D. 16
E. 18
OA is D-16... U all got it right!!
A certain roller coaster has 3 cars, and a passenger is equally likely to ride in any 1 of the 3 cars each time that passenger rides the roller coaster. If a certain passenger is to ride the roller coaster 3 times, what is the probability that the passenger will ride in each of the 3 cars?
A-0
B-1/9
C-2/9
D-1/3
E-1
Let the 3 cars on roller coasters be = A,B,C
on the 1st time, person can sit on any one of the cars, So probability = 1
on the 2nd time, person can sit one of the remaining 2 cars, so probability = 2/3
on the 3rd time, person can sit only of the remaining 1 cars, so probability = 1/3
Total probability = 1*(2/3)*(1/3)
= 2/9
Answer ( C )
A certain roller coaster has 3 cars, and a passenger is equally likely to ride in any 1 of the 3 cars each time that passenger rides the roller coaster. If a certain passenger is to ride the roller coaster 3 times, what is the probability that the passenger will ride in each of the 3 cars?
A-0
B-1/9
C-2/9
D-1/3
E-1
Its C..
Analysation: A person can sit in any of d cars in 3 ways for d 1st visit.
For d three visits, he can sit in 3*3*3 ways.
Now for d 1st visit , he can sit in 3c1 way.
2 nd visit, 2c1 way.
3rd visit, 1 way.
Prob =(3c1*2c1*1)/27=2/9
Hope its correct..
Puys .. plz clarify..Only integers can be even or odd??
For eg: 6^1/2 , 8^1/2( wat bout des no???)
( ^- sqr root)
thanks
Hi, Per my understanding fractions/decimals are also either even or odd because an integer can be represented as a fraction....
4/2 ~ 2 even
8^1/2 ~ 2.82 , so even
puys, any other ideas?
If N is a seven digit natural number such that when it is added to the sum of its own digits the result is the least eight digit number. What is the remainder when N is divided by 9?
OPTIONS
1)0
2)5
3)9
4)1
5)3