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Problem 10

# An experiment is given together with an event. Find the (modeled) probability of each event, assuming that the coins and dice are distinguishable and fair, and that what is observed are the faces or numbers uppermost. Two coins are tossed; the result is one or more heads.

Expert verified
The probability of getting one or more heads when tossing two fair and distinguishable coins is $$\frac{3}{4}$$ or 75%.
See the step by step solution

## Step 1: List all possible outcomes

When two coins are tossed, there are a total of 4 possible outcomes, as each coin has 2 faces, heads (H) and tails (T). The possible outcomes can be represented as: 1. Coin 1 shows heads and Coin 2 shows heads (HH) 2. Coin 1 shows heads and Coin 2 shows tails (HT) 3. Coin 1 shows tails and Coin 2 shows heads (TH) 4. Coin 1 shows tails and Coin 2 shows tails (TT)

## Step 2: Identify the outcomes for the given event

In this step, we will identify the outcomes from the list obtained in Step 1 that satisfy the given event: "one or more heads." The outcomes that match this condition are: 1. HH (Coin 1 shows heads and Coin 2 shows heads) 2. HT (Coin 1 shows heads and Coin 2 shows tails) 3. TH (Coin 1 shows tails and Coin 2 shows heads)

## Step 3: Calculate the probability

Now, we will calculate the probability of the event "one or more heads." To do this, we will divide the number of favorable outcomes by the total number of outcomes: $$P(\text{one or more heads}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$$ Since there are 4 possible outcomes when tossing two coins and 3 of them satisfy the event "one or more heads," the probability will be: $$P(\text{one or more heads}) = \frac{3}{4}$$ Therefore, the probability of getting one or more heads when tossing two fair and distinguishable coins is $$\frac{3}{4}$$ or 75%.

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