Today’s topic is:
This is one of my favourite topic of Mathematics because this is only single formula based Topic in early stage but afterwards in advance level probability may consist some advance topics like (permutation, combination, statistics…)
But in this tutorial I’ll be explaining only that single formula i.e.
P(E) = n(f) / n(t)
P(E) – probability of an event
n(f) – no of favourable outcome
n(t) – no of total outcome
event is refer to happening the task
outcomes are refer to result which can be favourable or non favourable too,
- this can be understood briefly with following example-
suppose we have a coin and we have to toss this now the result may vary with head or tail,but the possibility of occurring either head or tail are equally likely i.e. 50-50.
Now toss the same coin twice (or tossing two coins one time) the result may be in the forms
HH, HT, TT, TH
H – head T- tail
means both head can be resultant or bot tail can be or either head and tail or vice verse, In this situation we are having four outcomes.
If It is ask to find the probability of having head at least one time then favourable outcomes will be
HH, HT, TH
hence P(E)= 3/4
and if it is ask to find the probability of having at most one head then favourable outcomes will be
HT, HH, TH
again P(E)= 3/4.
One more thing I want to add in this thread is sum of the probabilities of all elementary events are equals to one(1).
If I pick first situation of having at least one head P(E) = 3/4
and probability of having no heads
outcome will be TT
P(E’) = 1/4
then sum of these probabilities of elementary events are
3/4 + 1/4 = 4/4
For more explanation watch following video
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