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A coin is tossed 10 times Find the probablity of obtaining head at least once

Probability of obtaining head at least once    =  1 - ( Probability of obtaining tail all 10 times)

= 1 - (1/2)^10

= 1023/1024

Therefore , probability is 1023/1024
selected
Sample space size $= 1024$
No. of favourable cases $= 1024-1=1023$
Probability $= \frac{1023}{1024} =99.90 \%$
answered by (870 points) 1 2 17
1023/1024

Here we can use Binomial distribution, for r success   = nCpr(1-p)n-r

here in Question mentioned that atleast one then it will go from 1 to 10.

Here we can direclty write for all come tail will be = 1 - (1/2)10

= 1023/1024.

+1 vote