Because of the 36 ways the two dice could fall, TWO satisfy the condition that you get a four and a six: die one has a four and die two has a six OR die one has a six and die two has a four. That's two chances out of 36, which simplifies to 1/18.
Quote:
|
2. A family has five kids. What is the probability that the first and last born are male? I have no clue how to get this. Obviously, the probability of getting a single male or female is 1/2, but I can't handle the order... What do I do? The answer is 1/4.
|
The kids in the middle are a red herring. The question really is: given two kids, what are the odds they're both male. Probabilities multiply when you're talking about a string of outcomes like this, so we're talking 1/2 * 1/2 = 1/4.
Quote:
|
3. What is the probability of flipping a coin six times in a row and getting three heads in a row, followed by three tails? What do I do with this? The answer's 1/64.
|
Again, a string of outcomes means you multiply the probabilities of each individual (in this case) flip. (1/2)^6 = 1/64.
You could do this longhand too, if you really wanted to get a tactile feel for this sort of problem. You could make a chart figuring out all of the possible orders in which six coin flips would come out. You'd find that there are 64 possible ways those six flips could turn out, of which only ONE would be "HHHTTT".
Quote:
|
4. What is the probability of flipping a coin six times in a row and getting three heads in a row and three tails? Once again, how do I account for the order? The answer's 5/16.
|
Not the way you've stated it here, it isn't. The way you've stated it, it's the same problem as #3, and the answer is 1/64. And I hope my answer on #3 served to back you down off the ledge about "accounting for the order". There's no need--multiply the odds of each individual element of the series, and you've got your odds for the series.