Answer:
* Here we assume that the coin tossed is fair.
Probability of event A = # Favorable cases for event A/ # Exhaustive cases for event B
Where
Favorable cases means outcomes of an experiment that favors a particular event
Exhaustive cases means, number of total possible outcomes.
For tossing a coin one time:
# Favorable cases = 1 (getting head)
# Exhaustive cases = 2 (two outcomes, say, head and tail)
So, P(getting head in a single toss) = 1/2
For tossing a coin two times:
# Favorable cases = 1 (getting 2 heads, say, HH)
# Exhaustive cases = 2*2 or 2^2 (4 outcomes, say, HH, HT, TH, TT)
So, P(getting 2 heads in two tosses of a coin) = 1/4
Similarly, for tossing a coin five times:
# Favorable cases = 1 (getting 3 heads, say, HHHHH)
# Exhaustive cases = 2*2*2*2*2 or 2^5 (32 outcomes, say, HHHHH, THHHH, ..... etc)
So, P(getting head in 5 tosses of a coin) = 1/32
Explanation: