Match the following lists:

Answer

`a to r , s ; b to p , q, r, s ; c to p , q, r, s ; d to p `

Explanation

`a to r,s, b to p,q,r,s, c to p , q, r, s, d top.`
a. P(success) = 1/2, P(failure)=1/2
Suppose 'n' bombs are to dropped. Let E be the event that the bridge is destroyed. Then, P(E) = 1 - P(0or 1 success)
`=1-(((1)/(2))^(n)+""^(n)C_(1)1/2((1)/(2))^(n-1))`
`=1-1/10ge(n+1)/(2^(n))or (2^(n))/(10(n+1))ge1`
b. The bag contains 2 red 3 white and 5 black balls. Hence `P(S)=1//5,P(F)=4//5,LetE` be the event of getting a red ball.
`P(E) =P(S or FSorFFS or ...]ge1/2`
The value of n consistent is 4.
c. Let there be x red socks and y blue socks and `x gt y.` Then
`(""^(x)C_(2)+""^(y)C_(2))/(""^(x+y)C_(2))=1/2`
`or (x(x-1)+y(y-1))/((x-y)(x+y-1))=1/2`
Multiplying both sides by `2(x+y)(x+y-1)` and expanding, we get
`2x^(2)-2x+2y^(2)-2y=x^(2)+2xy+y^(2)-x-y`
Rearranging, we have
`x^(2)-2xy+y^(2)=x+y`
or `(x-y)^(2) =x+y`
or ` |x-y|=x+y`
`Now, x+y le17`
`x-y lesqrt17`
As x- y must be an integer, so
`x-y=4`
`therefore x+y=16`
Adding both together and dividing by 2 yields `x ge 10.`
d. Let the number of green socks be `x gt 0.` Let E: be the event that two socks drawn are of the same colour.
`P(E)=P(R R or BB or WW or GG)`
`=(3)/(""^(6+x)C_(2))+(""^(x)C_(2))/(""^(6+x)C_(2))`
`=(6)/((x+6)(x+5))+(x(x-1))/((x+6)(x+5))=1/5`
`implies5(x^(x)-x+6)=x^(2)+11x+30`
`or 4x^(2)-16x=0`
`x=4`