Answer:
[tex]\sf 6)\quad \dfrac{3}{25}=12\%[/tex]
[tex]\sf 7) \quad \dfrac{1}{11}=9.1\%[/tex]
[tex]\sf 8) \quad \dfrac{7}{48}=14.6\%[/tex]
Step-by-step explanation:
[tex]\boxed{\sf Probability\:of\:an\:event\:occurring = \dfrac{Number\:of\:ways\:it\:can\:occur}{Total\:number\:of\:possible\:outcomes}}[/tex]
The events of the first spin landing on an odd number and the second spin landing on 2 are independent events because the outcome of the each spin does not affect the outcome of the other spin. Therefore, the probabilities remain the same for each spin.
Since the spinner has an equal chance of landing on each of its five numbered regions, we can determine the probability of the first spin landing on an odd number and the second spin landing on 2 by multiplying the probabilities of each individual event.
There are three odd numbers (1, 3, and 5), so the probability of the first spin landing on an odd number is:
[tex]\sf P(X=odd)=\dfrac{3}{5}[/tex]
There is only one region on the spinner with the number 2, therefore the probability of the second spin landing on 2 is:
[tex]\sf P(X=2)=\dfrac{1}{5}[/tex]
To find the probability of both events happening, multiply the individual probabilities:
[tex]\sf Probability=\dfrac{3}{5} \times \dfrac{1}{5}=\dfrac{3}{25}[/tex]
Therefore, the probability that the first spin lands on an odd number and the second spin lands on 2 is 3/25 or 12%.
[tex]\hrulefill[/tex]
When you choose a mystery novel for the first book, the number of mystery novels and the total number of remaining books change. This means that the probability of selecting a mystery novel for the second book is influenced by the previous selection. Therefore, the events are dependent because the outcome of the first book selection affects the probability of the second book selection.
To find the probability that both books chosen are mystery novels, we need to calculate the probability of selecting a mystery novel for the first book and then, given that the first book was a mystery novel, the probability of selecting another mystery novel for the second book.
There are a total of 4 mystery novels out of the 12 books available. So, the probability of selecting a mystery novel as the first book is:
[tex]\sf P(Mystery\;1)=\dfrac{4}{12}=\dfrac{1}{3}[/tex]
After selecting a mystery novel for the first book, there will be 3 mystery novels remaining out of the total 11 remaining books. Therefore, the probability of selecting another mystery novel for the second book, given that the first book was a mystery novel, is:
[tex]\sf P(Mystery\;2)=\dfrac{3}{11}[/tex]
To find the probability of both events happening, multiply the individual probabilities:
[tex]\sf Probability=\dfrac{1}{3} \times \dfrac{3}{11}=\dfrac{3}{33}=\dfrac{1}{11}[/tex]
Therefore, the probability of randomly selecting two mystery novels is 1/11 or approximately 9.1%.
[tex]\hrulefill[/tex]
The events in this scenario are dependent because the total number of shirts changes after the first selection, which affects the probability of the second selection.
To calculate the probability of wearing a blue shirt on Monday and a white shirt on Tuesday, we need to consider the number of blue shirts and white shirts, as well as the total number of shirts remaining after the first selection.
There are a total of 7 blue shirts out of the 16 shirts available. So, the probability of selecting a blue shirt on Monday is:
[tex]\sf P(Blue\;on\;Monday)=\dfrac{7}{16}[/tex]
After selecting a blue shirt on Monday, we have all 5 white shirts still available, but the total number of shirts is reduced to 15 (since one shirt has been chosen). Therefore, the probability of selecting a white shirt on Tuesday, given that a blue shirt was worn on Monday, is:
[tex]\sf P(White\;on\;Tuesday)=\dfrac{5}{15}=\dfrac{1}{3}[/tex]
To find the probability of both events happening, multiply the individual probabilities:
[tex]\sf Probability=\dfrac{7}{16} \times \dfrac{1}{3}=\dfrac{7}{48}[/tex]
Therefore, the probability of wearing a blue shirt on Monday and a white shirt on Tuesday is 7/48 or approximately 14.6%.
