Question

Champion bowler. A certain bowler can bowl a strike \(70 \%\) of the time. What's the probability that she a) goes three consecutive frames without a strike? b) makes her first strike in the third frame? c) has at least one strike in the first three frames? d) bowls a perfect game (12 consecutive strikes)?

Short answer

a) 0.027 b) 0.063 c) 0.973 d) 0.01384

Step-by-step solution

Understanding the Problem

The problem involves calculating probabilities for a bowler who can bowl a strike 70% of the time. We need to compute probabilities for different scenarios involving consecutive frames (like a bowling game). We identify probabilities for strikes and non-strikes from this information.

Define Basic Probabilities

Given the bowler's probability of striking is 70%, denoted as \( P(S) = 0.7 \). The probability of not striking (a non-strike) is the complement, so \( P(N) = 1 - P(S) = 0.3 \).

Probability of No Strikes in Three Frames (Part a)

We want the probability that the bowler does not strike in three consecutive frames. This means non-strikes in all three frames. Calculate \( P(N, N, N) = P(N)^3 = 0.3^3 \).

Calculate Probability for Part a

Compute \( 0.3^3 = 0.027 \). Therefore, the probability of going three consecutive frames without a strike is 0.027.

Probability of First Strike in Third Frame (Part b)

Here, the bowler must miss in the first two frames and strike in the third. This is computed as \( P(N, N, S) = P(N)^2 \times P(S) = 0.3^2 \times 0.7 \).

Calculate Probability for Part b

Compute \( 0.3^2 \times 0.7 = 0.063 \). Thus, the probability of making the first strike in the third frame is 0.063.

At Least One Strike in the First Three Frames (Part c)

To find the probability of at least one strike, use the complement rule: 1 minus the probability of no strikes in all three frames. Computed as \( 1 - P(N, N, N) = 1 - 0.3^3 \).

Calculate Probability for Part c

Compute \( 1 - 0.027 = 0.973 \). Therefore, the probability of having at least one strike in the first three frames is 0.973.

Probability of a Perfect Game (12 Strikes, Part d)

For a perfect game, the bowler must strike 12 consecutive times. This is calculated as \( P(S)^{12} = (0.7)^{12} \).

Calculate Probability for Part d

Compute \( (0.7)^{12} \approx 0.01384 \). Thus, the probability of bowling a perfect game is approximately 0.01384.

Key concepts

Complement Rule

In probability, the Complement Rule is a simple yet powerful concept. It states that the probability of an event not occurring is equal to 1 minus the probability that it does occur. This becomes especially useful when calculating probabilities like finding 'at least' events, which might be more complex to compute directly.
For example, if you're calculating the probability of a bowler having at least one strike in three frames, you can use the complement rule. Instead of calculating the probability of scoring 1, 2, or 3 strikes individually and summing them up, you calculate the complement: the probability of zero strikes, and then subtract from 1. If the probability of zero strikes in three frames is given by \( P(N, N, N) = 0.027 \), the complement rule tells us that \( P(\text{at least one strike}) = 1 - 0.027 = 0.973 \).
This approach simplifies calculations considerably and is applicable in many scenarios where dealing directly with multiple outcomes would otherwise be cumbersome.

Consecutive Events

When dealing with events occurring in sequence, such as consecutive strikes in bowling, understanding consecutive events is crucial. Consecutive events often require finding the probability that each event in a series happens continuously without interruption, which can denote success or failure.
In our example, the probability of three consecutive non-strikes helps us calculate the likelihood of the bowler failing to strike three times in a row. This is mathematically represented as the probability of a non-strike, raised to the power of how many times it occurs consecutively. Here, \( P(N)^3 = 0.3^3 = 0.027 \).
The probability of consecutive events multiplies the probabilities of individual events occurring in sequence. So the chances of a long string of successful events, like 12 strikes, quickly decrease with each additional event due to their consecutive nature.

Binomial Probability

Binomial probability is concerned with finding the probability of a certain number of successes in a fixed number of trials, with each trial having two possible outcomes, usually termed "success" and "failure."
The solutions to this exercise partly navigate this concept when considering independent trials across different frames of bowling, each with the same probability of success. For instance, in the scenario where the bowler needs to strike exactly for one of the first three frames, understanding binomial probability helps structure such calculations efficiently.
We can craft a formula for specific conditions, such as getting exactly one strike in a series of frames, by using the binomial probability formula: \(P(X = k) = \binom{n}{k}p^k(1-p)^{n-k} \), where \( n \) is the number of trials, \( k \) is the number of successes, \( p \) is the probability of success, and \( \binom{n}{k} \) is the combination function.

Independent Events

Independent events in probability are those whose outcomes do not influence each other. Essentially, the outcome of one event doesn't alter the probability of the other events occurring. This concept is crucial when calculating probabilities over multiple trials or sequences, such as consecutive bowling strikes or framing the question for the third-strike event.
Each frame is an independent event for the bowler. This means that whether or not she strikes in one frame doesn't affect whether she strikes in another. Mathematically, when events are independent, the probability of them all occurring is the product of their individual probabilities. This is why \( P(N, N, S) = P(N)^2 \times P(S) \) works.
Understanding independence ensures that calculations are based on correct assumptions, particularly in cases like predicting outcomes across games or trials.