Probability of tossing five coins and getting at least one head here is my dilemma. I want to know the probability of getting at least one head in five coins being tossed one after the other. Could you help me get the logic of this as it involves both mutually exclusive and independent events? Also, if possible please explain the logic with permutations and combinations. It would be much helpful.

Lenin
Hint: look at the complementary event, $A=\{\text{getting 5 tails}\}$. If you know how to compute $\mathbb{P}(A)$, then the probability you are looking for is $1-\mathbb{P}(A)$.



As an alternative (more complex) way of proving it, you can consider the 5 disjoint events $B_1,\dots,B_5$, where $B_i$ is defined as the event of getting exactly $i$ heads out of 5 (independent and fair) coin tosses. Then, observing that there are exactly $\binom{5}{i}$ sequences of $5$ draws containing $i$ heads (why?), and that each of them has probability exactly $p^i p^{5-i} = \frac{1}{2^5}$ (the last equality, as we assume $p=1/2$ since the coin is not biased), you get that the probability you are looking for is
$$
\mathbb{P}\left(\bigcup_{i=1}^5 B_i\right) =\sum_{i=1}^5 \mathbb{P}\left( B_i\right) = \frac{1}{2^5} \sum_{i=1}^5 \binom{5}{i} = \frac{1}{2^5}\left(\sum_{i=1}^5 \binom{5}{i} - \binom{5}{0} \right) = \frac{1}{2^5}\left(2^5 - 1 \right)
$$
where the first equality relies on the fact that the $B_i$'s are disjoint.
In an intuitive way, P(E)=$\frac{Favorable \ Outcomes}{Total \ Outcomes}$

So the to count the favorable outcomes: There are two possible outcomes for each toss. And there are 5 total tosses. So for each outcome there will be another outcome, so we'll multiply $2*2*2*2*2=2^5$

Sample Space or $n(S)=2^5$

Now think of getting one or more heads. We know that sum all probabilities is $1$. So we think of the other side. Why don't we calculate the number of possibilities for no head? And that would just be $\frac{1}{2}.\frac{1}{2}.\frac{1}{2}.\frac{1}{2}.\frac{1}{2}=\frac{1}{32}$

Probability of getting no head=$\frac{1}{32}$

So $P(E)=1-\frac{1}{32}$ = 
$\frac{31}{32}$

So this probability will be the required one.