Question

Two tennis professionals, \(A\) and \(B\), are scheduled to play a match; the winner is the first player to win three sets in a total that cannot exceed five sets. The event that \(A\) wins any one set is independent of the event that \(A\) wins any other, and the probability that \(A\) wins any one set is equal to .6. Let \(x\) equal the total number of sets in the match; that is, \(x=3,4,\) or \(5 .\) Find \(p(x)\).

Short answer

The probability distribution for x is: - p(3) = 0.280 - p(4) = 0.3312 - p(5) = 0.576

Step-by-step solution

Case 1: Total of 3 sets

In this case, either A wins all the sets or B wins all the sets. The probability of A winning all sets is: $$ p_A(3) = \binom{3}{3} (0.6)^3 (1-0.6)^{3-3} = 0.6^3 = 0.216 $$ The probability of B winning all sets is: $$ p_B(3) = \binom{3}{0} (0.6)^0 (1-0.6)^{3-0} = 0.4^3 = 0.064 $$ So, the total probability of having 3 sets is: $$ p(3) = p_A(3) + p_B(3) = 0.216 + 0.064 = 0.280 $$

Case 2: Total of 4 sets

In this case, one of them won 3 sets, and the other won 1 set. The probability of A winning 3 sets and B winning 1 set is: $$ p_A(4) = \binom{3}{2} (0.6)^3 (1-0.6)^{3-2} = 3 \times 0.6^3 \times 0.4 = 0.2592 $$ The probability of B winning 3 sets and A winning 1 set is: $$ p_B(4) = \binom{3}{1} (0.6)^1 (1-0.6)^{3-1} = 3 \times 0.6 \times 0.4^2 = 0.072 $$ So, the total probability of having 4 sets is: $$ p(4) = p_A(4) + p_B(4) = 0.2592 + 0.072 = 0.3312 $$

Case 3: Total of 5 sets

In this case, one of them won 3 sets, and the other won 2 sets. The probability of A winning 3 sets and B winning 2 sets is: $$ p_A(5) = \binom{4}{2} (0.6)^3 (1-0.6)^{4-2} = 6 \times 0.6^3 \times 0.4^1 = 0.3456 $$ The probability of B winning 3 sets and A winning 2 sets is: $$ p_B(5) = \binom{4}{2} (0.6)^2 (1-0.6)^{4-2} = 6 \times 0.6^2 \times 0.4^2 = 0.2304 $$ So, the total probability of having 5 sets is: $$ p(5) = p_A(5) + p_B(5) = 0.3456 + 0.2304 = 0.576 $$

Final Answer

Therefore, the probability distribution for x is: $$ p(3) = 0.280 \\ p(4) = 0.3312 \\ p(5) = 0.576 $$

Key concepts

Binomial Coefficients

Binomial coefficients play a crucial role in determining probabilities in situations where there are multiple instances of independent events. They are numbers that help us find out how many different ways we can pick a specific number of successes (e.g., winning sets) from a larger pool (e.g., total sets played). In math, they are represented with a formula and written as \( \binom{n}{k} \), meaning "n choose k". This is calculated using:

• \( n! \) is the factorial of n, meaning the product of all positive integers up to n.
• \( k! \) is the factorial of k.
• \( (n-k)! \) is the factorial of \( n-k \).

So, for example, using binomial coefficients, we found that player A winning 3 sets out of 3 is calculated as \( \binom{3}{3} \). This showcases how this formula helps break down the probability of different scenarios.

Independent Events

Independent events are occasions where one outcome does not affect another. In tennis, this means the result of each set is not influenced by previous ones. So even if player A wins every set, that doesn't raise or lower the chance of them winning the next one. For probability calculations, independence means we simply multiply the probabilities together. For example, if player A's chance to win a set is 0.6, it remains 0.6 for every set, regardless of past set results. Additionally, the probability of multiple independent events happening can also be combined by multiplication, like seeing how many times player A can win in a best-of-five series.

Probability Calculation

Calculating probabilities involves simple math once the groundwork is laid with concepts like binomial coefficients and understanding independent events. To predict how likely it is for player A or player B to win the series (in 3, 4, or 5 sets), we determine the chances of different sequences happening.When we say player A's probability of winning each set is 0.6, this drives the subsequent calculations. For example:

• Probability of A winning 3 sets and B winning none in 3 total sets: \( 0.6^3 = 0.216 \)
• This applies similarly across other scenarios with appropriate binomials and powers.

Understanding these calculations helps predict potential match outcomes. By adding up these possibilities, we form a probability distribution like in the step-by-step coding above, where the results such as \( p(3) = 0.280 \) are derived by weighting each possible outcome's chance.