How many outcomes (sequences of either head or tail) are possible if a coin with probability for head 1/3 and tail 2/3 is tossed 5 times?
32
The question asks about the total number of possible outcomes (sequences of heads or tails) when a coin is tossed 5 times. It mentions the probability of getting a head (1/3) and a tail (2/3), but this information about probability is not needed to determine the total number of possible sequences. The number of outcomes depends only on the number of possibilities for each individual toss and how many times the coin is tossed.
For a single toss of a coin, there are two possible outcomes:
These are the only two possibilities for each toss, regardless of the coin's probability bias.
When a coin is tossed multiple times, and each toss is an independent event, the total number of possible sequences is found by multiplying the number of outcomes for each toss together.
In this case, the coin is tossed 5 times. For each of the 5 tosses, there are 2 possible outcomes (Head or Tail).
To find the total number of possible sequences for 5 tosses, we multiply the number of outcomes for each toss:
Total Outcomes = (Outcomes per toss) × (Outcomes per toss) × (Outcomes per toss) × (Outcomes per toss) × (Outcomes per toss)
Total Outcomes = $2 \times 2 \times 2 \times 2 \times 2$
This can be written using exponents:
$$\text{Total Outcomes} = 2^5$$
Now, we calculate the value of $2^5$:
$$2^5 = 2 \times 2 \times 2 \times 2 \times 2 = 4 \times 2 \times 2 \times 2 = 8 \times 2 \times 2 = 16 \times 2 = 32$$
So, there are 32 possible outcomes or sequences when a coin is tossed 5 times.
The total number of possible sequences of outcomes (Heads or Tails) when a coin is tossed 5 times is 32. Each toss has 2 independent outcomes, and with 5 tosses, the total is $2^5$.
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