A biased coin is tossed until a head occurs. If the probability of heads on any given toss is .4, What is the probability that it will take 7 tosses until the first head occurs?
The answer i got was , (.60)^2(.40)
Now for the second part it says, what is the probability that it will take 9 tosses until the second head occurs.
Is the answer for this part be 9C2(.40)^2(.60)^7 or 8C1(.40)(.60)^5 I can't figure out if its 9 choose 2 or 8 choose 1... thank you
In the first part it should be (0.6)^6 * 0.4 because it says 7th toss is the first head. So, first 6 are tail and the 7th one is head. So, probability of that is (0.6)^6 * 0.4
In the second part it says 9 tosses until the second head. Means the 9th one is head we fixed that. In first 8 we got 1 head and rest tail.
So, it will be 8C1 * 0.4 * 0.6^7 * 0.4 = 0.035831808
0.6^7 because we need 7 tails, first 0.4 for the head coming in first 8 tosses and last 0.4 is to get head on the 9th toss.
I hope this will help.
Please comment if any doubt. Thank you.
A biased coin is tossed until a head occurs. If the probability of heads on any...
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