i have no idea how this works :/ e.g coins are tossed, what's the probability of getting atleast one head? in my revision book it gives me this formula which makes no sense: 1 - p(less than that many) if the formula dont help, explain to me in your way plz ^_^
btw i meant to say 3 coins omg but n works as well lol
So if, say, a fair coin is tossed three times, the probability of a head or tail will be equal (50%).
At least one head means that out of the three trials, you will get a head at least once. You therefore need to consider all possible outcomes: three heads, two heads and one head. That will be P(HnHnH)+P(HnHnT)+P(HnTnT). I'll assume you know the multiplication and addition rules.
think about it, the question is asking you to find the chance of getting atleast (which implies greater than or equal to) one head. assuming you toss the coin two times, you can get the following combinations: T T H H T H H T from inspection it's possible to deduce the probability of getting greater than or equal to one head (i.e. P(HH OR TH OR HT) = P(HH) + P(TH) + P(HT)). P(HH OR TH OR HT) can be given by 1 - P(TT) because the sum of probabilities must add up to 1 and TT is the only combination where there isn't a head. the same concept applies for n tosses but obviously there will be more combinations
If you toss n coins, can you work out the probability of getting less than one heads(bearing in mind that the number of heads has to be a whole number)?
probability of at least 1 head is 1-probability of all tails.
so for 3 coins tossed.
probability of 3 tails is 0.5x0.5x0.5=0.125
probability of at least one head is 1-0.125=0.875.
thanks, this is what eventually happened at the end of the revision guide but the fact that they worded the formula weirdly got me confused :/ yours was in context which is so much easier to understand