A fair coin is tossed 5 times. Compute the probability of tossing 5 heads in a row.

50%

Step-by-step explanation:

There is 2 side out of 100% half is 50 :D gl and mb if its wrong!

* 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, )

# Exhaustive cases = 2*2*2*2*2 or 2^5 (32 outcomes, say, , T, etc)

So, P(getting head in 5 tosses of a coin) = 1/32

Step-by-step explanation:

the function makes a strait line horizontally with no x intercept but a y intercept at 9.