# Probability q

1. Jul 30, 2009

### mathsgeek

I need help to answer this.

In a game of two up, twqo coins are tossed, bets are placed on whether coins give even (HH or TT) or odss (HT or TH). In the previous 8 games, evens has occured each time. What is the best bet on the ninth game?

This is what i did but i dont think tis right
P = 1/(2^8) (Probability of 8 evens) + 0.5 (Probability of 1 odd)

Also i have one other q.\\A blind person has 3 drawers of sock. Each draw contains a red, black and white sock. If a sock is taken from each draw:
what is the probability of two socks are black?
what is the probability that one sock is red?

For these 2, i calculated the number of ways= 3^3 = 27, but dont know how to go from here

Thanks

Last edited: Jul 30, 2009
2. Jul 30, 2009

### HallsofIvy

I have no idea why you think this is "Calculus and Analysis". I am moving it to "Set Theory, Logic, Probability, and Statistics".

3. Jul 30, 2009

### HallsofIvy

If we are to assume that each flip is independent of the others, even and odd are equally like to come up this time, just like all the times. One might, however, conclude that the coins are NOT fair and bet "even".

That makes no sense to me. By using "0.5" you are assuming that the coins are fair and the fact that it came up even 8 times before is irrelevant. In any case, you have not said what "P" is! I assume it is a probability but you didn't say probability of what nor does the question ask for a probability.

The probability of "Black, Black, Other color", in that order, is (1/3)(1/3)(2/3)= 2/27. There are three different orders for "two black socks" (the others are "Black, Other Color, Black" and "Other Color, Black, Black" so the probability of two black socks is (2/27)(3)= 2/9.

The probability of "Red, Other Color, Other Color", in that order, is (1/3)(2/3)(2/3)= 4/27. Again, there are 3 different order (the others are "Other Color, Red, Other Color" and "OTher Color, Other Color, Red) so the probability of exactly one sock red is (4/27)(3)= 4/9.

4. Jul 30, 2009

### mathsgeek

Thanks, With the socks is it basically P(B) x P(B) x P(other colour) x 3 (number of combinations per 2 Blacks) for the probability? Also, With the coin so all games are independent? however, if i was asked, whats the probability of getting evens 8 times in a row, would the P(8 evens)= 1/(2^8) ? And say, if we wanted probability of two heads, 8 times in a row, would it be P(2Hs)= 1/(4^8)? Thanks

I have one last q i need help with. "A box contains 10 calculators and there is 60% chance that it is defective. If a person chooses four calculators, what is the probability all 4 are defective? Thanks (show calculations using combinations or permutations where possible)

With this, i assumed there 6 defective calculators in the box, then used combinations to find a possible of 210 combinations. Then did 6c4 to find ways to choose 4 defective calculators and got 15. Therefore, 15/210 = 1/14, is this correct? Or do you go 6/10 x 5/9 x 4/7 x 3/6 = 71/210

Last edited: Jul 30, 2009
5. Aug 1, 2009

Any1?

6. Aug 2, 2009

### HallsofIvy

Surely that is not a direct quote! "A box contains 10 calculators and there is a 60% chance that it is defective". Grammatically that "it" must refer to the box and I have no idea how a box can be defective! I think you mean "A box contains 10 calculators and there is a 60% chance that any one is defective. That does NOT necessarily mean that there are 6 defective calculators in the box of 10 (that would be the "expected" value).

And if the problem did say "exactly 6 out of the 10 calculators in the box are defective" you woud find the probability that the first 4 you take out of the box are defective this way:

You take one calculator out of the box. There is a 6/10 chance it is defective.

If it is defective, then there are 5 defective calculator left in the box of 9 calculators. The probability that the next calculator is also defective is 5/9.

If it also is defective, then there are 4 defective calculators left in the box of 8 calculators. The probability that the second calculator you select is also defective is 4/8.

If it is defective, then there are 3 defective calculators left in the box of 7 calculators. The probability that the fourth calculator you select is also defective is 3/7.

I'll let you calculate the probability that the fourth calculator is defective. The probability all four calculators are defective is the product of all those probabilities.

But if you know that each calculator has a probability of .6 of being defective, then the probability of all four you choose being defective is just .64.