Expert: Scott A Wilson - 12/3/2009

Question
hi there. i couldn't get the write answer. a penalty shoot-out in a game of hockey requires each of 2 players to take a penalty hit to try to score a goal. in a simple model, each player has a probability of 0.8 of scoring a goal, and independence is assumed, calculate the probability that one goal is scored from the two hits.
in an alternative model, the probability of the second player scoring is reduced to 0.7 if the first player does not score. calculate the probability that the second player has scored, given that only one goal is scored.

2)a club social committee consists of eight people, two of whom are nicky and sam.two of the eight committee members are to be chosen at random to organize the next club disco.
by considering a tree diagram or otherwise.(the tree diagram is first choice having 3 people nicky(1/8),sam(1/8),other(6/8) and the second choice is for each person so nicky would have sam(1/7), other(6/7) and sam would have nicky(1/7), other(6/7) and other would have (nicky(1/7),sam(1/7),other(5/7)))
a) find the probability that both nicky and sam are chosen,given that at least one of nicky and sam is chosen?

3)trigonometry question is what does it mean when they give 4sinA+3cosA+7 and they ask find the least value as A varies?

and if they ask to find the exact value for cos2A,giving your answer as fraction i know the answer is cos^A=1/13 but how  did they get that  answer.

Answer
1) This means that the 2nd player scored and the 1st didn't.
Since the 1st player has a probability of 0.8 of scoring a goal,
that means his probability of not scoring is 0.2.
The second playerhas a probability of 0.7 of scoring.
The answer is found by multiply 0.2*0.7, or 2*7/100 = 14/100.

2) There is N (Nicky) S(Sam) and 6 other people.
Two members are chosen at random.
Let C(a,b) = a!/(b!(a-b)!)
The probability of N being chosen only is C(1,1)C(1,0)C(6,1)/C(8,2).
The probability of S being chosen only is C(1,0)C(1,1)C(6,1)/C(8,2).
The probability of both N and S being chosen is C(1,1)C(1,1)C(6,0)/C(8,2).
C(1,0) = 1!/(1!0!)=1, C(1,1)=1!/(0!1!)=1, C(6,1)=6!/5!=6, C(8,2)=8!/(2!6!)=8*7/(2*1)=56/2=28.
Probability of choosing N is then 1*1*6/28.
Probability of choosing S is then 1*1*6/28 as well.
Probability of choosing both is 1/28.

So the probabability of choosing either is the sum of the probability of choosing either
minus the probability of choosing both.  This is because they both include the chance of
choosing both, so we have to subtract it once.  It is 6/28 + 6/29 - 1/28 = 11/28.

So the probability of choosing both given that at least one is picked is
(1/28)/(13/28) = 1/13.

3) The way to minimize f(A) = 4sinA + 3cosA + 7 is to find the derivative.
It is f'(A) = 4cosA - 3sinA.  Setting 4cosA - 3sinA = 0 gives sinA/cosA = 4/3.
It is known that sinA/cosA = tanA.
So find the smallest angle that has tanA = 4/3, and that would be an angle in quadrant I.

For cos(2A), it is known that cos(2A) = 2cos²A - 1.
Given that cosA=1/13, cos(2A) = 2(1/13)²-1 = 2/169 - 169/169 = -167/169.