induction (with probability) (1 Viewer)

[Masaken]

Hugh toasted n different slices of bread: T1, T2, ..., Tn, and he buttered one of the sides for each slice. However, he accidentally dropped all of the slices of toast to the floor.

The different thicknesses of butter resulted in the probability of the kth slice, Tk, landing on its buttered side being equal to (1/2k+1) [2k+1 is the denominator, sorry I don't know how to use latex]

Use mathematical induction to show the probability for n pieces of buttered toast, that an odd number of slices will land on their buttered side is (n/2n+1) [again 2n+1 is the entire denominator]

did the base case and the assumption, but i have no idea how to prove that the probability is true for n = k+1 (P(k+1)), help please, thanks in advance

 

[HazzRat]

Hugh toasted n different slices of bread: T1, T2, ..., Tn, and he buttered one of the sides for each slice. However, he accidentally dropped all of the slices of toast to the floor.

The different thicknesses of butter resulted in the probability of the kth slice, Tk, landing on its buttered side being equal to (1/2k+1) [2k+1 is the denominator, sorry I don't know how to use latex]

Use mathematical induction to show the probability for n pieces of buttered toast, that an odd number of slices will land on their buttered side is (n/2n+1) [again 2n+1 is the entire denominator]

did the base case and the assumption, but i have no idea how to prove that the probability is true for n = k+1 (P(k+1)), help please, thanks in advance

Is this an actual maths question or is it a just a retelling of how my breakfast went this morning?

 

[Masaken]

Is this an actual maths question or is it a just a retelling of how my breakfast went this morning?

it is an actual math qn from an exercise I'm doing

 

[HazzRat]

rip. In yr 11 math we just learnt what a function is.

 

[jjoo8919]

@Masaken 3U math induction test this friday, but I decided to give this a shot. Do you still need help or have you got it figured out?

 

[jjoo8919]

[Inductive step only] To prove for n=k+1, P(odd number land on butter for (k+1)th toast) = (k+1)/[2(k+1)+1] = (k+1)/(2k+3)

From assumption: P(odd number land on butter for kth toast) = k/(2k+1)

P(odd number land on butter for (k+1)th toast) = k/(2k+1) + 1/(2k+3) * 1/(2k+1)
This is by adding the probabilities for the kth toast with the product of probability of kth toast landing buttered and (k+1)th toast landing buttered.

Since the assumption already accounts for all the possibilities up to k, multiplying the probability of kth toast landing buttered and (k+1)th toast landing buttered gives you the probability that the (k+1)th toast will............. help..

 

[unofficiallyred12]

uh this is a bit late since i just saw this, but my method is as follows for the inductive step.

First off, we split the k+1 pieces of bread falling into k pieces falling and then the last piece falling

Then we take two cases:

1- after the first k bread have fallen, the number of bread slices that land on buttered side is odd
By induction hypo, prob of this occurring is k/(2k+1). If we an odd number of slices landing on the buttered side to be odd when k+1 slices have fallen, then we need the (k+1)th piece to not land on the buttered side, prob of this = 1-1/(2k+3)= (2k+2)/(2k+3)
By multiplication principle, total probability of this case = k(2k+2)/(2k+1)(2K+3)

2- after the k bread have fallen, no. of bread slices that land on buttered side is even
by induction hypo, prob of this occurring = 1-k/(2k+1) = (k+1)/(2k+1)
Now we need the (k+1)th piece to fall on it's buttered side, prob of this = 1/(2k+3)
Total prob for this case = (k+1)/(2k+1)(2k+3)

Total prob = [(2k+2)+(k+1)]/(2K+1)(2k+3) = (2K+1)(k+1)/(2K+1)(2k+3) = (k+1)/(2k+3) as required