Expert: Josh - 5/14/2009

Question
A and B play chess every day. The probability that A beats B in any game is twice the probability that B beats A. The probability that B beats A is three times the probability that the game ends in a draw. What is the probability that any given game ends in a draw?

A. 3/5
B. 1/9
C. 1/3
D. 2/3
E. 1/5

Correct Answer: B

The probability that a child will get A, B, C or D grades is 0.4, 0.35, 0.15 and 0.10. Find the odds in favour of a child�s receiving at most C grade.

A. 1:5
B. 1:3
C. 2:5
D. 4:5
E. 2:3

Correct Answer:E

Four persons are chosen at random from a group of 3 men, 2 women and 4 children. Find the probability of selecting 1 man, 1 woman and 2 children.

A. 1/5
B. 2/5
C. 2/7
D. 5/13
E. 1/25

Correct Answer: C

I dont know how they got the answers.. I got different answers for all the questions.. thank you for your help..  

Answer
Hi Sara,

Brief summary:

Q1. The answer I got is 1/10. Refer to worked solution below.

Q2. P(C)+P(D) = 0.15+0.1 = 0.25
None of the options are viable.

Q3. Define combination C(N,k)=N!/[k!(N-k)!]

Number of ways to form a {M,W,C,C} group is P(4,2)=12.
- The set consists of MWCC, MCWC, MCCW, WMCC, WCMC, WCCM, CMWC, CMCW, CWMC, CWCM, CCMW, CCWM.

Probability of getting an arbitrary combination of {M,W,C,C} is
R=(3/9)*(2/8)*(4/7)*(3/6)=1/(3*2*7).

Probability of any {M,W,C,C} grouping is P(4,2)*R = 2/7

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More about Q1:

A game of chess has 3 possible outcomes: WIN (for A), LOSS (for A), or a DRAW. Use x, y, z to denote the probability of a WIN, LOSS and DRAW, respectively, for player A.

A WIN for player A automatically means a LOSS for player B (and vice versa). Using the information given, we translate "probability that A beats B in any game is twice the probability that B beats A" into "probability of a WIN for A is twice the probability of a WIN for B (this second possibility is same as a LOSS for A)".

The variables of interest are x and y. The word "twice" is incorporated in the equation x=2*y; it tells us "by how much". We will call this equation (1) so we can refer to this later.

x = 2y .....(1)

This equation may be interpreted as follows: The probability of a WIN for player A is twice the probability of a LOSS for A. Exactly what we were told.

Don't forget, we somehow have to get "z" involved. Here, z represents the probability of a DRAW. Using similar reasoning, we translate "probability that B beats A is three times the probability that the game ends in a draw" into

y = 3z .....(2)

Remember this: Given three unknowns (x, y and z) we need three independent pieces of information, in order to work out the values of all three variables (x, y and z). So far, we have two equations. To get the remaining equation, we use the fact that the three events "WIN", "LOSS" and "DRAW" make up the entire probability space. i.e., it covers all possible outcomes for the game. So, x+y+z=1 .....(3).

Our task is to work out the probability that a game ends in a DRAW, i.e., to find the value of "z". You may use any algebraic techniques that you are comfortable with to solve for z. Typically, we substitute one equation into another, to reduce the number of unknowns. This process gets us closer to finding a solution for "z". For instance, we may try to eliminate "x" first, thus making the problem of three unknowns (x, y and z) into a smaller problem that consists of two unknowns (y and z). And we keep going, try to use the information we have (do some arithmetic) to eliminate "y" from the remaining equation(s) to find a solution for "z".

You can also add/subtract one equation from another, without changing the nature of the relations. The key is to be CONSISTENT. Whatever you do (add, subtract, multiply etc.) to one side of the equation, you perform the same operation to the other side. There are no fast and hard rules as to what is best, we use whatever strategy is most convenient to us for a given situation.

With the three equations
x=2y    ...(1)
y=3z    ...(2)
x+y+z=1 ...(3)

we can subtract (1) from (3), to get
x+y+z - x = 1-2y  ....which simplifies to
y+z = 1-2y  ....Aim: express y in terms of z (first, add 2y to both sides)
3y+z = 1    ....then, subtract z from both sides
3y = 1-z    ....then, divide both sides by 3
y=(1-z)/3   ....(4)

Now, we have used up equation (1) and (3). What remains to be done is to plug (4) into (2). This gives
(1-z)/3 = 3z
Multiplying both sides by 3, we get
1-z = 9z
z=0.1

We can easily verify that y=0.3, x=0.6 and x+y+z=1. It looks like the model answer is wrong.