`10` different books and `2` different pens are given to `3` boys so that each gets equal number of things. The probability that the same boy does not receive both the pens is
A. `5//11`
B. `7//11`
C. `9//11`
D. `6//11`
Correct Answer - C
`(c )` `m(S)=(12!)/(4!4!4!3!)3!=(12!)/(4!4!4!)`
`n(E)=` the number of ways in which any one boy gets both the pens
`="^(3)C_(1)^(10)C_(2)(8!)/(4!4!2!)2!=3xx45(8!)/(4!4!)`
`:.P(E)=(3xx45(8!)/(4!4!))/((12!)/(4!4!4!))=(2)/(11)`
`:.` Required probability `=1-(2)/(11)=(9)/(11)`
Suppose than p and q, respectively, denote the probabilities that a thing goes to a man and a women.
`p=(a)/(a+b)and q=(b)/(a+b)`
Now, the probabilities of 0,1,2,3,… thing going to men...
Correct Answer - D
According to the givn condition,
`""^(n)C_(3)((1)/(2))^(n)=""^(n)C_(4)((1)/(2))^(n),`
where n is the number of times die is thrown.
`therefore""^(n)C_(3)=""^(n)C_(4)impliesn=7`
Thus, the required probability is
`=""^(7)C_(1)((1)/(2))^(7)=(7)/(2^(7))=(7)/(128)`
Correct Answer - C
The required probability is
1- probability of getting equal number of heads and tails
`=1-""^(2n)C_(n)((1)/(2))^(n)((1)/(2))^(2n-n)`
`=1-((2n)!)/((n!)^(2))xx(1)/(4^(n))`
Correct Answer - C
`(c )` `B_(1)B_(2)B_(3)…B_(7)B_(8)B_(9)B_(10)`
`(i)` When two terminal books are taken (`B_(1)B_(2)` or `B_(9)B_(10)`) then number of ways `=2xx7=14`
`(ii)` When two consecutive terminal books are not taken...
Correct Answer - A
`(a)` At event `A` he gets a number greater than `3`
`:.P(A)=1//2`
At event `B` he gets `5` in last throw
`:.P(B)=1//6`
`:.` Required probability is
`P(E)=(1)/(6)+(1)/(2)*(1)/(6)+(1)/(2)*(1)/(2)*(1)/(6)+…..oo`...