A fair coin is tossed 10 times. What is the probability that two tails occur consecutively?
Give your answer correct to 5 decimal places.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
May not be the most efficient way but here goes. First consider the number of ways you can toss a fair coin 2^10 = 1024. Now consider how many of those combinations do not involve consecutive tails
These are all the ways we can avoid consecutive tails:
With no tails there is 1 way this can happen all heads. = 1
With 1 tail there are 10 ways as the tail can be anywhere. = 10
With 2 tails we can imagine tails must occupy the 9 gaps between the 8 heads. We must choose 2 gaps from 9. 9C2 = 36
Again using 3 tails with 7 heads. We must choose 3 gaps from 8. 8C3 = 56
With 4 tails. We must choose 4 gaps from 7. 7C4 = 35
With 5 tails. We must choose 5 gaps from 6. 6C5 = 6
So no.ways we can avoid consecutive tails = (1+10+36+56+35+6)/(1024) = 9/64 = 0.140625
1-(0.140625) yields answer = 0.859375