Hi everybody and sorry about my English
I’ve a simple question and it seems typical but please before answering it thinks carefully because it is not simple as it appears

If we have a coin and we toss this coin 3 times, what is the probability of having a head one time or more (out of these three tosses)

Please explain to me why if you know the answer

I’ll try to rewrite the problem and make sure it’s clear
What is the probability of getting one head in any of the three tosses (in other words) getting one head or more in the three attempts?
I used this equation in excel =BINOMDIST(3,3,0.5,true) where the first 3 represents probability of getting 3 or fewer head , the other 3 represents the number of event and .5 represent the probability
The result was 1 !!
Who it can be one?! this is impossible
There is no such probability
One means we would have it for sure
But if we r going to run these trials, it is not guaranteed for us that we would have a head
It seems to me there is an integration formula by which the end result of such case is one, as I think it shouldn’t be one at all

In the BINOMDIST function, change "true" to "false" and subtract it from 1

Without using Excel, it is 1 - P(no heads) = 1 - (0.5)^3 = 1 - 0.125 = 0.875

Thank u JohnM for answering my question :wave:

I tried to solve another problem with same method
The probability of getting number 2 in flipping a dice (one time or more) if we flip a dice tow times
1-(1/6*1/6*5)= 0.861111111

Same idea as before, but with different probabilities.

= 1 - P(no 2's) = 1 - (5/6)^2 = 0.306

Thanks again JonhM :tup:
I think I need to go back to school and take an intensive course in stat :confused: