Free Math Worksheets

- What is the complement of a stage detecting the attempt? - What must happen at the first two stages for the third stage to be the first detector? - Which rule applies when independent events must occur in sequence?

1. Let \(p=0.80\) be the detection probability and \(q=1-p=0.20\) be the miss probability. 2. The first stage stops the attempt with probability \(p=0.80\). 3. For the third stage to be the first to detect the attempt, the first two stages must miss and the third must detect it. The probability is \(q^2p=(0.20)^2(0.80)=0.032\). 4. If no stage detects the attempt, all three stages miss. The probability is \(q^3=(0.20)^3=0.008\).

1. \(0.80\) 2. \(0.032\) 3. \(0.008\)

- Identify the exact sequence required for the first detection to occur on inspection \(3\). - Multiply the probabilities along that one sequence.

1. The required sequence is no defect, no defect, defect. 2. Its probability is \(P(X=3)=(0.7)^2\cdot 0.3=0.147\).

\(0.147=14.7\%\)

- For \(X=k\), identify the failures before the first success. - “At most three” includes success on spin \(1\), \(2\), or \(3\). - For part c, determine what must happen on the first four spins.

1. The success probability is \(p=\frac{1}{5}=0.20\), and the failure probability is \(0.80\). 2. \(P(X=2)=0.80\cdot 0.20=0.16\). 3. \(P(X=3)=(0.80)^2\cdot 0.20=0.128\). 4. \(P(X\le 3)=1-P(\text{no blue in the first three spins})=1-(0.80)^3=0.488\). 5. More than four spins are required exactly when the first four spins are not blue, so \(P(X>4)=(0.80)^4=0.4096\).

a) \(P(X=2)=0.16\); \(P(X=3)=0.128\) b) \(P(X\le 3)=0.488\) c) \(P(X>4)=0.4096\)

- Use the complement of finding at least one component needing rework. - Write an inequality involving the probability of no successes in \(n\) trials. - Test consecutive whole-number values near the threshold.

1. The probability that a component does not need rework is \(0.75\). 2. The probability of finding at least one component needing rework in \(n\) inspections is \(P(X\le n)=1-(0.75)^n\). 3. Require \(1-(0.75)^n\ge 0.80\), or \((0.75)^n\le 0.20\). 4. For \(n=5\), \(1-(0.75)^5\approx 0.7627\), which is less than \(0.80\). 5. For \(n=6\), \(1-(0.75)^6\approx 0.8220\), which meets the requirement. Therefore, the smallest value is \(n=6\).

\(n=6\)

- List the possible numbers of emails the customer can receive. - A fifth email is sent whenever the first four are not opened. - For fewer than five emails, use the probability that the first opening occurs on that email. - Multiply each possible value by its probability and add.

1. Let \(X\) be the number of emails sent. Then \(X\in\{1,2,3,4,5\}\), and the probability an email is not opened is \(q=0.88\). 2. The probabilities are \(P(X=1)=0.12\), \(P(X=2)=0.88(0.12)=0.1056\), \(P(X=3)=0.88^2(0.12)=0.092928\), \(P(X=4)=0.88^3(0.12)=0.08177664\), and \(P(X=5)=0.88^4=0.59969536\). The last probability includes every case in which the first four emails are not opened. 3. Therefore, \(E(X)=1(0.12)+2(0.1056)+3(0.092928)+4(0.08177664)+5(0.59969536)=3.93556736\).

The customer receives an average of about \(3.94\) emails.

- List the possible numbers of spins in one series. - The fourth spin occurs whenever the first three spins are losses. - Find the expected number of spins first. - Multiply the expected number of spins by the cost per spin.

1. Let \(X\) be the number of spins. Then \(X\in\{1,2,3,4\}\), with success probability \(p=0.25\) and failure probability \(q=0.75\). 2. The probabilities are \(P(X=1)=0.25\), \(P(X=2)=0.75(0.25)=0.1875\), \(P(X=3)=0.75^2(0.25)=0.140625\), and \(P(X=4)=0.75^3=0.421875\). 3. Thus, \(E(X)=1(0.25)+2(0.1875)+3(0.140625)+4(0.421875)=2.734375\) spins. 4. The expected cost is \(\$1.50\cdot2.734375=\$4.1015625\), which rounds to \(\$4.10\).

The expected total cost is about \(\$4.10\).