A coin is so biased that the heads occurs four times as frequently as tails. Another coin is biased such that the heads occurs 65% of the times. When the two coins are tossed simultaneously, what is the probability of at least one tail turning up?

Options

A. A)`35%`
B. B)`87%`
C. C)`48%`
D. D)`73%`

Answer

To solve the problem step by step, we need to determine the probability of getting at least one tail when tossing two biased coins simultaneously.

### Step 1: Determine the probabilities for Coin 1 (C1)
The first coin is biased such that heads occur four times as frequently as tails. This means:
- Let the probability of tails (T) be \( P(T) = x \).
- Then the probability of heads (H) will be \( P(H) = 4x \).

Since the total probability must equal 1:
\[ P(H) + P(T) = 1 \implies 4x + x = 1 \implies 5x = 1 \implies x = \frac{1}{5} \]
Thus, the probabilities for Coin 1 are:
- \( P(T) = \frac{1}{5} \)
- \( P(H) = \frac{4}{5} \)

### Step 2: Determine the probabilities for Coin 2 (C2)
The second coin is biased such that heads occur 65% of the time. Thus:
- \( P(H) = 0.65 \)
- \( P(T) = 1 - P(H) = 1 - 0.65 = 0.35 \)

So, the probabilities for Coin 2 are:
- \( P(T) = 0.35 \)
- \( P(H) = 0.65 \)

### Step 3: Calculate the probability of getting at least one tail
To find the probability of getting at least one tail when tossing both coins, it is easier to first calculate the probability of getting no tails (i.e., getting heads from both coins) and then subtracting this from 1.

The probability of getting heads from both coins is:
\[ P(H_1 \text{ and } H_2) = P(H_1) \times P(H_2) = \left(\frac{4}{5}\right) \times (0.65) \]
Calculating this gives:
\[ P(H_1 \text{ and } H_2) = \frac{4}{5} \times 0.65 = \frac{4 \times 0.65}{5} = \frac{2.6}{5} = 0.52 \]

### Step 4: Calculate the probability of at least one tail
Now, we can find the probability of getting at least one tail:
\[ P(\text{at least one tail}) = 1 - P(H_1 \text{ and } H_2) = 1 - 0.52 = 0.48 \]

### Step 5: Convert to percentage
To express the probability as a percentage:
\[ P(\text{at least one tail}) = 0.48 \times 100 = 48\% \]

### Final Answer
The probability of getting at least one tail when tossing the two coins simultaneously is **48%**.

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