Not sure if anybody has solved it already ... Good sum ..
we have 2 independent values ...rest all are dependent ... Let 1st term be x and 2nd term = y
Series is as follows :
x y y/x 1/x 1/y x/y x y . . . Hence....cyclicity is 6 ...every 7th term is same as 1st , 8th same as 2nd ans so on ....
So product of set of 6 nos is 1 .. Product of first 50 nos is same as product of first 2 nos (6*8 + 2) and product of all is same as product of first 4 terms (16*6 + 4)
Hence, 2 eqns are xy =27 and y^2/x = 27 .... Solving, we get x = 3 and y=9
simple one... A credit card number has 6 digits (between 1 to 9). The first two digits are 12 in that order, the third digit is bigger than 6, the forth is divisible by 3 and the fifth digit is 3 times the sixth. How many different credit card numbers exist?a)27.b)36.c)72.d)112.e)422.
Not sure if this approach is correct....
Possibilities for 3rd digit - 3 Possibilities for 4th digit - 3 Possibilities for 6th digit - 4
5th digit will always by 3 times 6th digit, so possibility - 1
Poss credit cards -> 3*3*4 = 36
IMO b) 36...
A small error in bold i believe ... We just have 3 poss for 6th digit since only for 1,2 and 3 is three times the no a single digit
One more for practice: In a blue jar there are red, white and green balls. The probability of drawing a red ball is 1/5. The probability of drawing a red ball, returning it, and then drawing a white ball is 1/10. What is the probability of drawing a white ball? a)1/5. b). c)1/3. d)3/10. e).
Solved in a hurry ...nt sure if it as straightforward
Selection of red and white balls is independent .. For 2 independent events A and B P(A and B) = P(A) * p(B)
simple one... A credit card number has 6 digits (between 1 to 9). The first two digits are 12 in that order, the third digit is bigger than 6, the forth is divisible by 3 and the fifth digit is 3 times the sixth. How many different credit card numbers exist?a)27.b)36.c)72.d)112.e)422.
A small error in bold i believe ... We just have 3 poss for 6th digit since only for 1,2 and 3 is three times the no a single digit
Adding one more: What is the probability for a family with three children to have a boy and two girls, assuming the probability of having a boy or a girl is equal? a)1/8 b) c) d)3/8 e)5/8
@ Bhavin : I am 100% sure ,you can easily solve these questions... :thumbsup: But I want others to try. Please Please Please post your answers and approach after some folks try... I just want all Puy's to attempt on probability sums...
If Sum is 8:- Lets say first Dice got 1, then possibilities for rest 2 dice are:- (1,6); (2,5); (3,4); (4,3); (5,2); (6;1) ==> 6 Lets say first Dice got 2, then possibilities for rest 2 dice are:- (1,5); (2,4); (3,3); (4,2); (5,1) ==> 5 Lets say first Dice got 3, then possibilities for rest 2 dice are:- (1,4); (2,3);(3,2); (4,1) ==> 4 Lets say first Dice got 4, then possibilities for rest 2 dice are:- (1,3); (2,2); (3;1) ==> 3 IIIy, fixing first dice as 5,6 ==> 2,1 possibilies for rest dices
Adding when sum is 8 => 6+5+4+3+2+1 = 21
IIy for sum 14 => 1+2+3+4+5 => 15
Probability = (21 + 15) / 6*6*6 => 1/6 Answer is A
One more: What is the probability of getting a sum of 8 or 14 when rolling 3 dice simultaneously? a)1/6 b) c) d)21/216 e)32/216
Total possibilities: 2*2*2 = 8 1 B & 2 G possibility = (G,G,B); (G,B,G); (B,G,G) = 3
So, Ans: 3/8 (D)
Adding one more: What is the probability for a family with three children to have a boy and two girls, assuming the probability of having a boy or a girl is equal? a)1/8 b) c) d)3/8 e)5/8
Case 1 : Consider 1st couple to sit together. (Hence they are considered as 1 unit) . So, 1 unit and remaining 3 people i.e total 4 people can be arrange in 4! ways and the couple themselves can be arranged in 2! ways.. Total ways in which 1st couple sit together = 2!*4! ways..
Case 2 : Consider 2nd couple to sit together: Similar to above calculation, Total ways in which 2nd couple sit together = 2!*4! ways..
Now, its like a venn diagram, where each of the above cases has 1 special case where both couples could be seated together and this special case has been included in each of the cases above , hence we need to subtract this special case once.
Special Case : (both couples together ) Now, when both couples are seated next to each other, Total no of ways = 2!*2!*3! ( each couple considered as 1 unit , and 2! ways for each of the couple to be seated among themselves )
Total ways (atleast 1 couple seated next to each other) = case 1 + case 2 - special case
If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?
If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?
A. 20
B. 40
C. 50
D. 80
E. 120
IMO : D
Total Number of ways in which 3 can be selected out of 10: 10C3 = 120 Number of ways in which 1 couple is chosen out of 5, and 1 individual from the rest: 5C1 * 8C1 = 40 Number of ways in which a couple is not chosen to be on committee is :120 -40 = 80
only one person can be selected from family, Ways of selecting 3 people = 5C3 each of 3 families have 2 members, So, Total ways = 5C3 *2*2*2 => 80
Ans (D)
Could someone explain with solution?
If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?
@ pj02 Though ur answer is correct, but i think the explaination is wrong. Ur last line says "Now these 3 can be arranged in 3! Ways." We are just asked to select people and not arrange. Please correct me if I am wrong.
-Nipun Bansal
One more way to solve this: 1st person can be selected in 10 ways. 2nd in 8 ways. 3rd in 6 ways
so total10*8*6=480 ways. Now these 3 can be arranged in 3! Ways.
Right triangle LMN is to be constructed in the xy-plane so that the right angle is at point L and LM is parallel to the x-axis. The x- and y- coordinates of L, M, and N are to be integers that satisfy the inequalities -3 (A) 72 (B) 576 (C) 4032 (D) 4608 (E) 6336
if values of x include both -3 and 4 and values of y include both 3 and 11...then L and M can have 8*7=56 x values and L and M can have 9*8 possible y values which in total gives (8*7)*(9* values=4032 values of (x,y)
Adding one more: What is the probability for a family with three children to have a boy and two girls, assuming the probability of having a boy or a girl is equal? a)1/8 b) c) d)3/8 e)5/8
Please post OA. I think order/arrangement here is not important, as the Probability of either of them(boy or gal) being born is same i.e 1/2 => 1/2*1/2*1/2=1/8. Option A
values=4032 values of (x,y)