If you toss a fair coin and you toss until a head is followed by a tail. What is the probability that at least 5 tosses are needed for this to occur?
you are looking for a consecutive H-T pair
possible pairs are
... H-H, H-T, T-H, T-T
there are 4 consecutive pairs in 5 tosses
the probability of H-T is .25 for any pair of tosses, so the probability of NOT H-T is .75
the probability of 3 consecutive NOT H-T pairs is ... .75^3
there are 32 possible outcomes (2^5) for 5 consecutive tosses
... you can list them to check the answer
When I made a tree diagram for 4 tosses of a coin... I had 6 branches where you tossed until a head was followed by a tail. That was 6 branches out of 16. Then I added another toss and came up with one more for the sample set. HHHHT, so that would be 1 out of 32 produce the head then tail. So the P(at least 5 flips) is 1 - P(less than 5 flips)
1 - (6/16 + 1/32)
I think I am on the right track. Please advise.
Or... do we not break up where the tree ended (as in my 1- solution), but do 1 - 7/32 which would then be 25/32 for the probability of it taking atleast 5 times for a head followed by a tail to occur?
This is a bit more complicated than it first appears
Let A be the event of ... HT
case 1. it happens in 2 tosses
HT**
Prob(A on 2 tosses) = 1(1/4)
case 2. (3 Tosses)
THT, HHT,
Prob(A on 3 tosses) = 2(1/8) = 1/4
case 3. (4 tosses)
TTHT, THHT, HHHT
prob(A on 4 tosses) = 3(1/16) = 3/16
Of course prob(A for an infinite number of tosses) = 1
so Prob( A as stated) = 1 - (1/4+1/4+3/16) = 5/16
Look over my cases and make sure I got them all
e.g. in case 3, we can't count outcomes like HTHT, since we would no longer toss after HT, so I did not list them.
The "atleast 5 tosses" part means 1-P(50rless tosses) so I think we also need the fifth toss that has 1 in 32 ways to get HT, that being TTTHT
So that has to go into the subtracting bracket as well... I think : ) Please advise.
"at least 5 tosses" to means:
5 tosses, or 6 tosses, or 7 tosses, etc
so we don't want 2, 3, or 4 tosses.
agree?
btw, 5 tosses would be quite a few more than TTTHT.
What about HHHHT, THHHT, TTTHT, etc
HHHHH ... 1/32
HHHHT ... 1/32
HHHT ... 1/16
HHT ... 1/8
HT ... 1/4
TTTTT ... 1/32
TTTTH ... 1/32
TTTHH , TTTHT ... 1/32 , 1/32
TTHHH , TTHHT , TTHT ... 1/32 , 1/32, 1/16
THHHH , THHHT , THHT , THT
... 1/32 , 1/32 , 1/16 , 1/8
these are the 16 ways to toss a coin until a head is followed by a tail, or 5 tosses are reached; along with their probabilities
the probability that an H-T pair occurs with FEWER than 5 tosses is ... 11/16
this agrees with Reiny's result
sorry about my 1st effort
Awesome! Now I truly understand the scenerio. Thanks so much for your wise and timely replies and solutions : )
To calculate the probability that at least 5 tosses are needed for a head to be followed by a tail, we can consider the possible scenarios where this condition is met.
Let's imagine the potential outcomes for each toss as H (for heads) and T (for tails). We need to find the probability of getting at least four consecutive tails (TTTT) followed by HT.
Here are the possible sequences of tosses that satisfy the condition:
1. TTTTH
2. TTTTTH
3. TTTTTTH
4. TTTTTTTH
5. TTTTTTTTH
6. ... (and so on)
As we can see, the number of consecutive tails required before getting HT is increasing by one for each new sequence.
Since we are dealing with a fair coin, the probability of getting a head or a tail in each toss is 0.5 (or 1/2).
The probability of getting at least 5 tosses can be calculated as follows:
P(at least 5 tosses) = P(4 consecutive tails) + P(5 consecutive tails) + P(6 consecutive tails) + ...
To find these individual probabilities, we multiply the probability of getting tails (0.5) for the number of consecutive tails we desire.
P(4 consecutive tails) = (0.5)^4
P(5 consecutive tails) = (0.5)^5
P(6 consecutive tails) = (0.5)^6
And so on.
To calculate the overall probability, we sum up these individual probabilities:
P(at least 5 tosses) = (0.5)^4 + (0.5)^5 + (0.5)^6 + ...
This is a geometric series with a common ratio of 0.5. We can use the formula for the sum of an infinite geometric series to find the result:
P(at least 5 tosses) = (0.5)^4 / (1 - 0.5)
Simplifying the equation, we have:
P(at least 5 tosses) = (1/16) / (1/2)
P(at least 5 tosses) = 1/8
Therefore, the probability that at least 5 tosses are needed for a head to be followed by a tail is 1/8, or 0.125, which is approximately 12.5%.