ProbabilityHard
Question
5 girls and 10 boys sit at random in a row having 15 chairs numbered as 1 to 15, then the probability that end seats are occupied by the girls and between any two girls an odd number of boys sit is:
Options
A.
B.
C.
D.
Solution
n(S) = ways of sitting of 10 boys and 5 girls = 15!
Let end seats are occupied by the girls & between first and second girl x boys are seated similarly between second and third y boys ............. so on then
x + y + z + w = 10
where x, y, z, w are (2k + 1) type
2k1 + 1 + 2k2 + 1 + 2k3 + 1 + 2k4 + 1 = 10
⇒ k1 + k2 + k3 + k4 = 3 where ki ≥ 0
number of solution are 3 + 4 - 1C4 - 1 = 6C3
n (E) = 6C3 ≠ 10! × 5!
Let end seats are occupied by the girls & between first and second girl x boys are seated similarly between second and third y boys ............. so on then
x + y + z + w = 10
where x, y, z, w are (2k + 1) type
2k1 + 1 + 2k2 + 1 + 2k3 + 1 + 2k4 + 1 = 10
⇒ k1 + k2 + k3 + k4 = 3 where ki ≥ 0
number of solution are 3 + 4 - 1C4 - 1 = 6C3
n (E) = 6C3 ≠ 10! × 5!
Create a free account to view solution
View Solution FreeMore Probability Questions
If 12 identical balls are to be placed in 3 identical boxes, then the probability that one of the boxes contains exactly...For 10 observations $x_{1},x_{2},\ldots.,x_{10}$, if $\sum_{i = 1}^{10}\mspace{2mu}\left( x_{i} + 2 \right)^{2} = 180$ a...For solving a problem, odds against to A are 4 : 3 and odds in favour to B are 7: 5. The probability that the problem wi...For two given events A and B, the relation P (AB) = P (A) P(B) implies that A and B are-...If a, b and c are three numbers (not necessarily different) chosen randomly and with replacement from the set {1, 2, 3, ...