Notes
\(\text{Probability}=\displaystyle \frac{\text{Successful Events}}{\text{Total Events}}\)
\(\text{Probability of 2 Independent Events}\)
\(\text{(Probability event 1)} \times \text{(Probability event 2)}\)
Practice Questions
\(\textbf{1)}\) There is a box of marbles. There are 12 red, 7 green and 11 blue marbles. You select 2 marbles.
What is the probability of both marbles being green? The answer is \(\frac{7}{30} \cdot \frac{6}{29}= \frac{42}{870}=\frac{21}{435}\approx 4.83\% \)
\(\textbf{2)}\) There is a box of marbles. There are 12 red, 7 green and 11 blue marbles. You select 2 marbles.
What is the probability of both marbles being blue? The answer is \(\frac{11}{30} \cdot \frac{10}{29}=\frac{110}{870}=\frac{11}{87}\approx 12.6\% \)
\(\textbf{3)}\) There is a box of marbles. There are 12 red, 7 green and 11 blue marbles. You select 2 marbles.
What is the probability of neither marble being red? The answer is \(\frac{18}{30} \cdot \frac{17}{29}=\frac{306}{870}=\frac{153}{435} \approx35.2\% \)
\(\textbf{4)}\) There is a box of marbles. There are 12 red, 7 green and 11 blue marbles. You select 2 marbles.
What is the probability of neither marble being green? The answer is \(\frac{23}{30} \cdot \frac{22}{29}=\frac{506}{870}=\frac{253}{435} \approx58.2\% \)
\(\textbf{5)}\) There is a box of marbles. There are 12 red, 7 green and 11 blue marbles. You select 2 marbles.
What is the probability of the first marble being red and the second blue without replacement? The answer is \(\frac{12}{30} \cdot \frac{11}{29}=\frac{132}{870}=\frac{22}{145} \approx 15.2\% \)
\(\textbf{6)}\) There is a box of marbles. There are 12 red, 7 green and 11 blue marbles. You select 2 marbles.
What is the probability of the first marble being blue and the second red without replacement? The answer is \(\frac{11}{30} \cdot \frac{12}{29}=\frac{132}{870}=\frac{22}{145} \approx 15.2\% \)
\(\textbf{7)}\) There is a box of marbles. There are 12 red, 7 green and 11 blue marbles. You select 2 marbles.
What is the probability of the first marble being red and the second blue with replacement? The answer is \(\frac{12}{30} \cdot \frac{11}{30}=\frac{132}{900}=\frac{22}{150} \approx 14.7\% \)
\(\textbf{8)}\) There is a box of marbles. There are 12 red, 7 green and 11 blue marbles. You select 2 marbles.
What is the probability of the first marble being blue and the second red with replacement? The answer is \(\frac{11}{30} \cdot \frac{12}{29}=\frac{132}{900}=\frac{22}{150} \approx 14.7\% \)
\(\textbf{9)}\) There is a box of marbles. There are 12 red, 7 green and 11 blue marbles. You select 2 marbles.
What is the probability of the first marble being green and the second blue with replacement? The answer is \(\frac{7}{30} \cdot \frac{11}{30}=\frac{77}{900} \approx 8.56\% \)
\(\textbf{10)}\) There is a box of marbles. There are 12 red, 7 green and 11 blue marbles. You select 2 marbles.
What is the probability of the first marble being green and the second blue without replacement? The answer is \(\frac{7}{30} \cdot \frac{11}{29}=\frac{77}{870} \approx 8.85\% \)
\(\textbf{11)}\) You flip a coin and roll a six-sided die.
What is the probability of getting a head and a 5? The answer is \(\frac{1}{2} \cdot \frac{1}{6}=\frac{1}{12}\approx 8.33 \% \)
\(\textbf{12)}\) You flip a coin and roll a six-sided die.
What is the probability of getting a tail and a 3? The answer is \(\frac{1}{2} \cdot \frac{1}{6}=\frac{1}{12}\approx 8.33 \% \)
\(\textbf{13)}\) You flip a coin and roll a six-sided die.
What is the probability of getting a tail and a 1? The answer is \(\frac{1}{2} \cdot \frac{1}{6}=\frac{1}{12}\approx 8.33 \% \)
\(\textbf{14)}\) You flip a coin and roll a six-sided die.
What is the probability of getting a tail and an even on the die? The answer is \(\frac{1}{2} \cdot \frac{1}{2}=\frac{1}{4}= 25 \% \)
\(\textbf{15)}\) You flip a coin and roll a six-sided die.
What is the probability of getting a tail and an odd on the die? The answer is \(\frac{1}{2} \cdot \frac{1}{2}=\frac{1}{4}= 25 \% \)