In each game of chess that Bobby plays against Boris there is a probability of 1/3 that Bobby wins, a prob. of 1/6 that Boris wins, and prob. of 1/2 that the game is drawn. They play 4 games.
i) find prob. that Bobby wins 2 games and Boris wins 2 games
ii) find prob. that Bobby wins 1 game, Boris wins 1 game and the other 2 games are drawn.
answer:
i) 4C2(1/3)^2(1/6)^2
ii)4C2(1/2)^2(1/3)(1/6) *2
i get the answer for i) but i dont understand why there is a need to multiply 2 for part ii)
Hey! So the answer without that extra multiplication accounts for selecting which games are drawn/won, then the probability of these occurring. We multiply by two to account for the fact that the two games won by Boris and Bobby can be swapped, as in, once we've chosen which games they'll win, there are then two different allocations to those games
ques2 :
Consider x^2=4ay with focus S. The normal at P(2ap, ap^2) meets y axis at R and triangle SPR is equilateral
i) normal at P is x+py=2ap+ap^3
ii) R(0, a(2+p^2)
iii) prove SP is equal in length to latus rectum, that is 4a units
not sure how to do part iii)
For Part (iii), we need to use the fact that the triangle SPR is equilateral, meaning SP=PR=RS. So, if we can prove any of those sides is equal to 4a units, our job is done. Let's just get expressions for all three (if you spot what to do straight away, that's awesome too):
All of these
must be equal by definition, therefore we can say that:
Then we go back and pop that in our distance equation for SP:
Hope this helps!