Question

1. a pet store has two tanks of tropical fish. the first tank contains 7 blue fish and 3 yellow fish and the second tank contains 10 blue fish and 8 yellow fish. if keisha randomly purchases a fish from each tank, what is the probability that both fish will be blue?

a. \\(\frac{17}{180}\\) b. \\(\frac{1}{21}\\) c. \\(\frac{7}{18}\\) d. \\(\frac{17}{28}\\)

1. suppose q and r are independent events, and \\(p(q) = 0.39\\), \\(p(r) = 0.85\\). find \\(p(q\text{ and }r)\\).

a. 1.24 b. 0.3315 c. 0.46 d. 0.1794

Explanation:

Question 7

Step1: Calculate total fish in first tank

First tank: \(7\) blue + \(3\) yellow = \(10\) fish. Probability of blue from first tank: \(\frac{7}{10}\).

Step2: Calculate total fish in second tank

Second tank: \(10\) blue + \(8\) yellow = \(18\) fish. Probability of blue from second tank: \(\frac{10}{18}=\frac{5}{9}\).

Step3: Multiply probabilities (independent events)

Probability both blue: \(\frac{7}{10} \times \frac{5}{9} = \frac{7 \times 5}{10 \times 9} = \frac{35}{90} = \frac{7}{18}\). Wait, but let's re - check. Wait, second tank: \(10\) blue and \(8\) yellow, total \(18\)? Wait, \(10 + 8 = 18\), yes. First tank: \(7+3 = 10\). So probability blue from first: \(\frac{7}{10}\), blue from second: \(\frac{10}{18}=\frac{5}{9}\). Then \(\frac{7}{10}\times\frac{5}{9}=\frac{7\times5}{10\times9}=\frac{35}{90}=\frac{7}{18}\). But the options: Option C is \(\frac{7}{18}\)? Wait, but let's check the options again. Wait, maybe I misread the second tank. Wait, the problem says "the second tank contains 10 blue fish and 8 yellow fish"? Wait, no, maybe it's a typo? Wait, no, let's re - calculate. Wait, first tank: 7 blue, 3 yellow (total 10). Second tank: 10 blue, 8 yellow (total 18). So probability blue from first: 7/10, blue from second: 10/18. Multiply: (7/10)*(10/18)=7/18. So the answer should be C. \(\frac{7}{18}\).

Question 8

Step1: Recall formula for independent events

For independent events \(Q\) and \(R\), \(P(Q\cap R)=P(Q)\times P(R)\).

Step2: Substitute values

Given \(P(Q) = 0.39\) and \(P(R)=0.85\). Then \(P(Q\cap R)=0.39\times0.85 = 0.3315\).

Question 7 Answer:

C. \(\frac{7}{18}\)

Question 8 Answer:

B. \(0.3315\)

Answer:

Step1: Recall formula for independent events

For independent events \(Q\) and \(R\), \(P(Q\cap R)=P(Q)\times P(R)\).

Step2: Substitute values

Given \(P(Q) = 0.39\) and \(P(R)=0.85\). Then \(P(Q\cap R)=0.39\times0.85 = 0.3315\).

Question 7 Answer:

C. \(\frac{7}{18}\)

Question 8 Answer:

B. \(0.3315\)