A fair coin is tossed 6 times. What is the probability of tossing at least 4 heads in a row?

that would be P(4H)+P(5H)+P(6H)

= 1/6^4 (1 + 1/6 + 1/36)

To calculate the probability of tossing at least 4 heads in a row when a fair coin is tossed 6 times, we need to look at the different possible outcomes.

First, let's determine the total number of possible outcomes when tossing a fair coin 6 times. For each toss, there are 2 possible outcomes: heads or tails. Since we toss the coin 6 times, the total number of possible outcomes is 2^6 = 64.

Next, we need to determine the number of outcomes where we get at least 4 heads in a row. In order to have at least 4 heads in a row, we can consider two scenarios: either the first 4 tosses are heads, or the last 4 tosses are heads.

1. First 4 tosses are heads: In this scenario, the last 2 tosses can be either heads or tails. So, there are 2 possibilities for the last 2 tosses. Therefore, the number of outcomes where the first 4 tosses are heads is 2.

2. Last 4 tosses are heads: In this scenario, the first 2 tosses can be either heads or tails. So, there are 2 possibilities for the first 2 tosses. Therefore, the number of outcomes where the last 4 tosses are heads is also 2.

Since these scenarios are mutually exclusive (they can't occur simultaneously), we can add up the number of outcomes for each scenario to get the total number of outcomes where we have at least 4 heads in a row. In this case, it is 2 + 2 = 4.

Finally, we can calculate the probability by dividing the number of favorable outcomes (4) by the total number of possible outcomes (64):

Probability = Number of favorable outcomes / Total number of possible outcomes
= 4 / 64
= 1/16

Therefore, the probability of tossing at least 4 heads in a row when a fair coin is tossed 6 times is 1/16.