all 1 = 1
all 2 = 1
one 2 six 1 = 7
two 2 four 1 = 15
three 2 two 1 = 10
total 34
In a box, there are some red balls, some blue balls and two yellow balls. The probability that a ball picked at random is red is 1/2. Also, the probability that two balls picked simultaneously
are blue is 1/11.How many balls are present in the box?
are blue is 1/11.How many balls are present in the box?
1) 22
2) 18
3) 12
3) 12
4) 16
5) 20
5) 20
Q. For every positive even integer n, the function h(n) is defined to be the product of all even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) +1, then p is
A. b/n 2 and 10
B. b/n 10 and 20
C. b/n 20 and 30
D. b/n 30 and 40
E. Greater than 40
p.s. b/n is between
B. b/n 10 and 20
C. b/n 20 and 30
D. b/n 30 and 40
E. Greater than 40
p.s. b/n is between
@soumitrabengeri said:In a box, there are some red balls, some blue balls and two yellow balls. The probability that a ball picked at random is red is 1/2. Also, the probability that two balls picked simultaneouslyare blue is 1/11.How many balls are present in the box?1) 22 2) 183) 12 4) 165) 20
12
@soumitrabengeri said:In a box, there are some red balls, some blue balls and two yellow balls. The probability that a ball picked at random is red is 1/2. Also, the probability that two balls picked simultaneouslyare blue is 1/11.How many balls are present in the box?1) 22 2) 183) 12 4) 165) 20
12?????
@soumitrabengeri said:In a box, there are some red balls, some blue balls and two yellow balls. The probability that a ball picked at random is red is 1/2. Also, the probability that two balls picked simultaneouslyare blue is 1/11.How many balls are present in the box?1) 22 2) 183) 12 4) 165) 20
red balls = n
blue balls = n - 2
yellow = 2
C(n - 2, 2)/C(2n, 2) = 1/11
(n - 2)(n - 3)/{2n(2n - 1)} = 1/11
Solving this we will get n = 6
So, 12 balls are there
@chillfactor said:f(n) is the number if ways of reaching nth stair.If first he climbs 1 step, then rest can be climbed in f(n - 1) waysIf first he climbs 2 step, then rest can be climbed in f(n - 2) waysSo, f(n) = f(n - 1) + f(n - 2)Now, f(1) = 1 and f(2) = 2Using them find f(8)f(3) = f(2) + f(1) = 3f(4) = 5f(5) = 8f(6) = 13f(7) = 21f(8) = 34So, 34 sequences are possible
SPAM - sry
sir, aap fibonacci uncle ko jante the kya
? aapke kafi solutions mien fibonacci series ka use dekha hai
sir, aap fibonacci uncle ko jante the kya
? aapke kafi solutions mien fibonacci series ka use dekha hai @soumitrabengeri said:In a box, there are some red balls, some blue balls and two yellow balls. The probability that a ball picked at random is red is 1/2. Also, the probability that two balls picked simultaneouslyare blue is 1/11.How many balls are present in the box?1) 22 2) 183) 12 4) 165) 20
12?
@soumitrabengeri said:In a box, there are some red balls, some blue balls and two yellow balls. The probability that a ball picked at random is red is 1/2. Also, the probability that two balls picked simultaneouslyare blue is 1/11.How many balls are present in the box?1) 22 2) 183) 12 4) 165) 20
12?
@soumitrabengeri said:In a box, there are some red balls, some blue balls and two yellow balls. The probability that a ball picked at random is red is 1/2. Also, the probability that two balls picked simultaneouslyare blue is 1/11.How many balls are present in the box?1) 22 2) 183) 12 4) 165) 20
12?
@Hypertexter said:Q. For every positive even integer n, the function h(n) is defined to be the product of all even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) +1, then p isA. b/n 2 and 10B. b/n 10 and 20C. b/n 20 and 30D. b/n 30 and 40E. Greater than 40p.s. b/n is between
greater than 40????
@soumitrabengeri said:In a box, there are some red balls, some blue balls and two yellow balls. The probability that a ball picked at random is red is 1/2. Also, the probability that two balls picked simultaneouslyare blue is 1/11.How many balls are present in the box?1) 22 2) 183) 12 4) 165) 20
3) 12? R=6;B=4;Y=2
@soumitrabengeri said:In a box, there are some red balls, some blue balls and two yellow balls. The probability that a ball picked at random is red is 1/2. Also, the probability that two balls picked simultaneouslyare blue is 1/11.How many balls are present in the box?1) 22 2) 183) 12 4) 165) 20
12
@soumitrabengeri said:In a box, there are some red balls, some blue balls and two yellow balls. The probability that a ball picked at random is red is 1/2. Also, the probability that two balls picked simultaneouslyare blue is 1/11.How many balls are present in the box?1) 22 2) 183) 12 4) 165) 20
red - r
total = 2r
blue = r-2
(r-2)/2r*(r-3)/(2r-1)=1/11
=>11(r^2-5r+6)=4r^2-2r
=>7r^2-53r+66=0
=>r=6
total=12
total = 2r
blue = r-2
(r-2)/2r*(r-3)/(2r-1)=1/11
=>11(r^2-5r+6)=4r^2-2r
=>7r^2-53r+66=0
=>r=6
total=12
@soumitrabengeri said:In a box, there are some red balls, some blue balls and two yellow balls. The probability that a ball picked at random is red is 1/2. Also, the probability that two balls picked simultaneouslyare blue is 1/11.How many balls are present in the box?1) 22 2) 183) 12 4) 165) 20
12?
@krum said:red - rtotal = 2rblue = r-2(r-2)/2r*(r-3)/(2r-1)=1/11=>11(r^2-5r+6)=4r^2-2r=>7r^2-53r+66=0=>r=6total=12
how did u get b=r-2 ??