Chapter 10, Problem 8RE

Solution

(a)

To calculate: The probability that the first strike in a game comes in the third frame.

The probability that the first strike in a game comes in the third frame.is 0.144.

Given information:

The bowlers make a strike on 40% of the frames.

P(S) = 40% = 0.4

Formula used:

Use the complement rule:

  $P\left(A^c\right)=P\left(notA\right)=1-P\left(A\right)$   ...... (1)

Calculation:

Let S = strike

The bowlers make a strike on 40% of the frames.

P(S) = 40% = 0.4

Substitute 0.4 for $P\left(S\right)$ in equation (1)

  $\begin{array}{c}P\left(\text{not}S\right)=1-P\left(S\right)\\ =1-0.4\\ =0.6\end{array}$

The first strike occurs in the third frame when two frames did not result in a strike and the 3^rd frame resulted in a strike.

P (First strike in the third frame) = P(not S)×P(not S)×P(S)    ...... (2)

Substitute 0.4 for $P\left(S\right)$ and 0.6 for P(not S) in equation (2)

  $\begin{array}{c}P\left(\text{First strike in the third frame}\right)=0.6\times 0.6\times 0.4\\ =0.144\end{array}$

Hence, the probability that the first strike in a game comes in the third frame.is 0.144.

(b)

To calculate: The probability that she makes a strike in at least one of the first 5 frames.

The probability that she makes a strike in at least one of the first 5 frames.is 0.922.

Given information:

The bowlers make a strike on 40% of the frames.

P(S) = 40% = 0.4

Formula used:

Definition of binomial probability:

  $P\left(X=k\right)=\left(\begin{array}{l}5\\ k\end{array}\right).p^k.{\left(1-p\right)}^{n-k}$   ...... (1)

Calculation:

n= number of trials=5

p=probability of success=40%=0.40

The number of successes among a fixed number of independent trials at a constant

probability of success follows a binomial distribution:

Substitute 0.4 for $p$ ,0 for $k$ and 5 for $n$ in equation (1)

  $\begin{array}{c}P\left(X=0\right)=\left(\begin{array}{l}5\\ 0\end{array}\right)\cdot {0.40}^0\cdot {\left(1-0.40\right)}^{5-0}\\ =\frac{5!}{0!\left(5-0\right)!}\cdot {0.40}^0\cdot {0.60}^5\\ =1\cdot 1\cdot {0.60}^6\\ =0.07776\end{array}$

Use the complement rule:

  $P\left(A^c\right)=P\left(notA\right)=1-P\left(A\right)$   ...... (1)

Substitute 0.07776 for $P\left(X=0\right)$ in equation (1)

  $\begin{array}{c}P\left(X\ge 1\right)=1-P\left(X=0\right)\\ =1-0.07776\\ =0.92224\end{array}$

Hence, the probability that she makes a strike in at least one of the first 5 frames.is 0.922

(c)

To calculate: The probability that she bowls an entire game (10 frames) without a strike.

The probability that she bowls an entire game (10 frames) without a strike is 0.0060.

The bowlers make a strike on 40% of the frames.

P(S) = 40% = 0.4

Formula used:

Definition of binomial probability:

  $P\left(X=k\right)=\left(\begin{array}{l}5\\ k\end{array}\right).p^k.{\left(1-p\right)}^{n-k}$   ...... (1)

Calculation:

n= number of trials=5

p=probability of success=40%=0.40

The number of successes among a fixed number of independent trials at a constant

probability of success follows a binomial distribution:

Substitute 0.4 for $p$ ,0 for $k$ and 10 for $n$ in equation (1)

  $\begin{array}{c}P\left(X=0\right)=\left(\begin{array}{l}10\\ 0\end{array}\right)\cdot {0.40}^0\cdot {\left(1-0.40\right)}^{10-0}\\ =\frac{10!}{0!\left(10-0\right)!}\cdot {0.40}^0\cdot {0.60}^{10}\\ =1\cdot 1\cdot {0.60}^{10}\\ \approx 0.0060\end{array}$

Hence, the probability that she bowls an entire game (10 frames) without a strike is 0.0060.