Understanding the Biased Coin Probability Problem

The problem asks for the probability of getting at most two tails in three independent tosses of a biased coin. First, we need to figure out the probabilities of getting a head (H) or a tail (T) with this biased coin.

Calculating Probabilities for a Biased Coin

We are told that heads comes up thrice as likely as tails. Let \(P(H)\) be the probability of getting a head and \(P(T)\) be the probability of getting a tail.

• Given: \(P(H) = 3 \times P(T)\).
• We also know that the sum of probabilities for all possible outcomes is 1: \(P(H) + P(T) = 1\).

Substitute the first equation into the second:

\(3 \times P(T) + P(T) = 1\)

\(4 \times P(T) = 1\)

\(P(T) = \frac{1}{4}\)

Now we can find \(P(H)\):

\(P(H) = 3 \times P(T) = 3 \times \frac{1}{4} = \frac{3}{4}\)

So, the probability of getting a tail is \(\frac{1}{4}\) and the probability of getting a head is \(\frac{3}{4}\).

Probability of "At Most Two Tails"

We are looking for the probability of getting "at most two tails" in three independent tosses. This means the number of tails can be 0, 1, or 2. The possible outcomes for three tosses are HHH, HHT, HTH, THH, HTT, THT, TTH, TTT.

Calculating the probability for each of these cases (0, 1, or 2 tails) and summing them up would be a bit lengthy. A simpler approach is to use the complementary event.

Using the Complementary Event

The complementary event to "at most two tails" (meaning 0, 1, or 2 tails) is "exactly three tails". If we find the probability of getting exactly three tails, we can subtract it from 1 to get the probability of getting at most two tails.

Let \(A\) be the event of getting at most two tails. Let \(A^c\) be the complementary event, getting exactly three tails.

\(P(A) = 1 - P(A^c)\)

Calculating Probability of Exactly Three Tails

Getting exactly three tails in three tosses means the outcome is TTT. Since the tosses are independent, the probability of this sequence is the product of the probabilities of each individual toss being a tail.

\(P(\text{TTT}) = P(T) \times P(T) \times P(T)\)

\(P(\text{TTT}) = \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4}\)

\(P(\text{TTT}) = \frac{1 \times 1 \times 1}{4 \times 4 \times 4} = \frac{1}{64}\)

So, the probability of getting exactly three tails is \(\frac{1}{64}\).

Calculating Probability of At Most Two Tails

Now, we can find the probability of getting at most two tails:

\(P(\text{At most two tails}) = 1 - P(\text{Exactly three tails})\)

\(P(\text{At most two tails}) = 1 - \frac{1}{64}\)

To subtract, find a common denominator:

\(1 = \frac{64}{64}\)

\(P(\text{At most two tails}) = \frac{64}{64} - \frac{1}{64} = \frac{64 - 1}{64} = \frac{63}{64}\)

Converting to Decimal

The probability is \(\frac{63}{64}\). To compare this with the options, we convert the fraction to a decimal:

\(\frac{63}{64} \approx 0.984375\)

Rounding to two decimal places, this is approximately 0.98.

Summing probabilities for 0, 1, 2 tails: \(\frac{27}{64} + \frac{27}{64} + \frac{9}{64} = \frac{27+27+9}{64} = \frac{63}{64}\).

Final Probability Result

The probability of getting at most two tails in three independent tosses of the biased coin is \(\frac{63}{64}\), which is approximately 0.98.

Probability Revision Table

Additional Probability Information

Understanding probability requires identifying outcomes, events, and the sample space. For multiple independent trials, like coin tosses, we often use concepts like the binomial distribution, although for a small number of trials like three, direct calculation or using the complementary event is also straightforward.

• Sample Space: The set of all possible outcomes. For three tosses, it's {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}. There are \(2^3 = 8\) possible outcomes.
• Event: A subset of the sample space. "At most two tails" is the subset {HHH, HHT, HTH, THH, HTT, THT, TTH}.
• Binomial Probability: For \(n\) independent trials, each with two outcomes (success/failure), where \(P(\text{success}) = p\) and \(P(\text{failure}) = 1-p\), the probability of getting exactly \(k\) successes is given by the formula: \(P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}\). In this problem, we could define 'tail' as a success with \(p=1/4\).
  • \(P(\text{0 tails}) = \binom{3}{0} (\frac{1}{4})^0 (\frac{3}{4})^3 = 1 \times 1 \times \frac{27}{64} = \frac{27}{64}\)
  • \(P(\text{1 tail}) = \binom{3}{1} (\frac{1}{4})^1 (\frac{3}{4})^2 = 3 \times \frac{1}{4} \times \frac{9}{16} = \frac{27}{64}\)
  • \(P(\text{2 tails}) = \binom{3}{2} (\frac{1}{4})^2 (\frac{3}{4})^1 = 3 \times \frac{1}{16} \times \frac{3}{4} = \frac{9}{64}\)
  • \(P(\text{3 tails}) = \binom{3}{3} (\frac{1}{4})^3 (\frac{3}{4})^0 = 1 \times \frac{1}{64} \times 1 = \frac{1}{64}\)
• The sum of probabilities for 0, 1, or 2 tails is \(\frac{27}{64} + \frac{27}{64} + \frac{9}{64} = \frac{63}{64}\), confirming the complementary method result.