A coin is loaded, so that the probability of a head occuring on a single toss is 2/3. In six tosses of the coin, what is the probability of getting all heads or all tails?

Well, I must say, that's one "loaded" question! If the probability of a head occurring on a single toss is 2/3, then the probability of a tail would be 1/3.

To calculate the probability of getting all heads or all tails in six tosses, we just need to multiply the probability of getting a head (2/3) or a tail (1/3) by itself six times.

So, the probability of getting all heads is (2/3)^6, which is approximately 0.13.

The probability of getting all tails is (1/3)^6, which is approximately 0.0046.

But don't be too disappointed if you don't get either of these outcomes. Remember, probability is just a fancy way of saying "chance." So, in the end, it's all just a coin toss!

To find the probability of getting all heads or all tails in six tosses of a loaded coin, we need to calculate the probability of each scenario separately and then add them together.

Probability of getting all heads:
The probability of getting a head on a single toss of the loaded coin is 2/3. Since we have six tosses, we need to multiply this probability by itself six times:
(2/3) * (2/3) * (2/3) * (2/3) * (2/3) * (2/3) = (2/3)^6

Probability of getting all tails:
The probability of getting a tail on a single toss of the coin is 1 - 2/3 = 1/3. Again, since we have six tosses, we need to multiply this probability by itself six times:
(1/3) * (1/3) * (1/3) * (1/3) * (1/3) * (1/3) = (1/3)^6

Now, we add the two probabilities together:
(2/3)^6 + (1/3)^6 = 64/729 + 1/729 = 65/729

Therefore, the probability of getting all heads or all tails in six tosses of the loaded coin is 65/729.

To find the probability of getting all heads or all tails in six tosses of a loaded coin, we can use the concept of binomial distribution.

The probability of getting a head on a single toss is given as 2/3, which means the probability of getting a tail is 1 - 2/3 = 1/3.

Let's calculate the probability of getting all heads first:
In six tosses, the probability of getting a head on each toss is (2/3) * (2/3) * (2/3) * (2/3) * (2/3) * (2/3) = (2/3)^6.

Similarly, the probability of getting all tails is (1/3) * (1/3) * (1/3) * (1/3) * (1/3) * (1/3) = (1/3)^6.

Since getting all heads or all tails are mutually exclusive events, we can add these probabilities together:
(2/3)^6 + (1/3)^6 = (64/729) + (1/729) = 65/729.

Therefore, the probability of getting all heads or all tails in six tosses of the loaded coin is 65/729.

The probability of all events occurring is found by multiplying the probabilities of the individual events.

Heads = (2/3)^6

Tails = (1/3)^6