probability?

Question

1. If the number of permutatuions of n objects taken r a a time is times the number of combinations of n objects taken r at a time, find the numerical value of r.
2. The probability of getting exactly k heads in 35 tosses of a fair coin is the same as the probability of getting exactly k+1 heads in 36 tosses. Find k.

Answer 1

1. nPr = r! * nCr. You must be missing part of the question.

2. The answer is k = 17. The number of ways to get exactly k heads in 35 tosses is 35!/[k!*(35-k!)]. The number of ways to get exactly k+1 heads in 36 tosses is 36!/[(k+1)!*(36-(k+1))!].

The probability of getting exactly k heads in 35 tosses is: {35!/[k!*(35-k)!)]} / 2^35.

The probability of getting exactly k+1 heads in 36 tosses is:
{36!/[(k+1)!*(36-(k+1))!]} / 2^36.

Let's set these two probabilities equal to each other:
{35!/[k!*(35-k)!)]} / 2^35 = {36!/[(k+1)!*(35 - k)!]} / 2^36

Now, multiply both sides by (35-k)!:
(35!/k!) / 2^35 = [36!/(k+1)!] / 2^36

Now, multiply both sides by k!:
35! / 2^35 = [36!/(k+1)] / 2^36

Now, multiply both sides by 2^35:
35! = [36!/(k+1)] / 2

Now, divide both sides by 35!:
1 = [36/(k+1)] / 2

Multiply both sides by 2:
2 = 36/(k+1)
k+1 = 36/2
k+1 = 18
k = 17

Therefore, the probability of getting 17 heads in 35 tosses is the same as getting 18 heads in 36 tosses. FYI, that probability is: 0.1320605996

Answer 2

1. Are you missing part of the decription here? You said "If the number ... is [?] times the number".

2. To caclulate the probability of getting exactly k heads in 35 tosses, you have to consider the fact that the 35 heads can turn up in a number of different ways. There are k! / 35! (35-k)! ways to arrange a series of 35 heads and (35-k) tails, and the probability that for a given sequence, the exact order of heads and tails shows up is (1/2)^k. So the probability is ((1/2)^k) k! / 35! (35-k)!.
The probability of getting k+1 heads in 36 tosses is likewise
((1/2)^k) 36! / (k+1)! (36-k-1)!

Setting them both equal to each other, you get
((1/2)^k) k! / 35! (35-k)! = ((1/2)^k) 36! / (k+1)! (36-k-1)!
k! / 35! (35-k)! = 36! / (k+1)! (36-k-1)!
k! (k+1)! (36-k-1)! = 36! 35! (35-k)!
k! (k+1)! (35-k)! = 36! 35! (35-k)!
k! (k+1)! = 36! 35!
(k+1) (k! k!) = 36 (35! 35!)
Looks like k = 35.

This doesn't seem right though, because the probability of getting 35 heads in 35 tosses is (1/2)^35, but getting 36 in 36 tosses is (1/2)^36. I suppose there are the trivial solutions where k > 35, and thus the probability is zero.

Answer 3

2. You can solve with symmetry considerations.

In order to have P(35) = P(36) then E(X) = k*p = 35.5, therefore k= 71.

Using my TI-83:
P(35 in 71 trials) = P( 36 in 71 trials) = 0.0937

+ add
I see that I solved a different problem. I found the number of tosses needed so that P(35) = P(36). Notice that symmetry could also have solved the correct problem:
E(X) = n*p = 35/2 = 17.5 this is the symmetry point so that P(17) = P(18).