probability

Question

(1 pt) Consider the experiment where a fair coin is tossed.
If the toss results in a head, then 1 die is thrown.
If the toss results in a tail, then 2 dice are thrown.
Let X denote the random variable that
counts the number of spots showing on the thrown die or dice.
The range of values that X can assume are the
positive integers from 1 to 12 inclusive.
Please give the corresponding probabilities for the values of X given below.
Pr(X = 1) =

Pr(X = 2) =

Pr(X = 3) =

Pr(X = 4) =

Pr(X = 5) =

Pr(X = 6) =

Pr(X = 7) =

Pr(X = 8) =

Pr(X = 9) =

Pr(X = 10) =

Pr(X = 11) =

Pr(X = 12) =
Further, find the probability that X is divisible by 5.
Probability that X is divisible by 5 equals

Answer 1

Well, the complexity of this problem isn't too bad, though there's a decent amount of work involved.

 

Consider the probability that X = 1.  This can only happen if one die is thrown, since two dice will always show at least two spots.  The probability of X being equal to 1 is therefore contingent on the coin toss turning up heads (50% chance) times the probability that the die will turn up a 1 (1/6 chance) - these have a product of 1/12, or .0833, so this is the probability of X = 1.

 

X = 2 is more complicated, since it can happen one of two ways.  If the coin turns up heads, the die can roll a 2; alternately, if the coin turns up tails, the two dice can turn up with two 1s.  The probability of X = 2 is therefore the sum of these two possibilities:

 

1/2 * 1/6, as before

and 1/2 * 1/36; there's 1 chance in 36 that two dice will both roll 1s (simply 1/6 * 1/6).

 

for a total probability of .0972

 

X=3 is similar to 2, save for the second part of the sum.  Instead of 1 in chance 36 of a 2 on two dice, you have 2 chances in 36, since you can roll a two on the first die and a one on the second, or vice versa.  The total probability is then

1/2 * 1/6 + 1/2 * 2/36 = .1111

 

X=4 is again similar, but with 3 chances in 36 instead of 2, since you can roll (1+3), (2+2), or (3+1) to satisfy the condition.  I'll leave it to you to work out the exact probability.

 

X=5 and 6 progress in the same way.

 

X = 7 is the first one again substantially different, since it's no longer possible to roll it on a single die; it's only possible to achieve if the initial coin toss turns up tails.  If this happens, there's 6 possible rolls of two dice that can yield this sum; (1+6),(2+5),(3+4),(4+3),(5+2), and (6+1).  The probability is then

1/2 * 6/36 = .08333

 

8 through 12 are then very similar to 7, except we now have decreasing numbers of ways to achieve the roll on two dice.  8 can only be rolled in 5 different ways; (2+6),(3+5),(4+4),(5+3),(6+2).  Nine can only be rolled in 4 ways, ten in 3, and so on, down to X = 12 only having a probability of 1/2 * 1/36 = .0139.

 

The probability that X is divisible by 5 can be answered easily once you have the full list of probabilities; it's divisible by 5 if X = 5 or 10, so the probability that it's divisible by five is just Pr(X=5) + Pr(X=10).