A coin is tossed 99 times.
What is the probability of getting all heads? Express your answer as a simplified fraction or a decimal rounded to four decimal places.
each toss is an independent event, so
Prob(99 consecutive heads)
= (1/2)^99 or appr 1.5777 x 10^-30
rather slim, I would say
To find the probability of getting all heads when a coin is tossed 99 times, we need to divide the number of favorable outcomes (getting all heads) by the total number of possible outcomes.
In this case, the coin has two possible outcomes - heads or tails. Since we want to get all heads, there is only one favorable outcome.
So, the probability of getting all heads is 1 out of 2^99 (two raised to the power of 99), which is:
1 / (2^99) = 1 / 5,764,607,523,034,234,710,536,640,000
This probability is an extremely small number. If we round it to four decimal places, it becomes:
0.0000
Therefore, the probability of getting all heads when a coin is tossed 99 times is approximately 0.0000.