The probability of selecting a dime or Canadian currency is 8/11
Further explanation
The probability of an event is defined as the possibility of an event occurring against sample space.
[tex]\large { \boxed {P(A) = \frac{\text{Number of Favorable Outcomes to A}}{\text {Total Number of Outcomes}} } }[/tex]
Permutation ( Arrangement )
Permutation is the number of ways to arrange objects.
[tex]\large {\boxed {^nP_r = \frac{n!}{(n - r)!} } }[/tex]
Combination ( Selection )
Combination is the number of ways to select objects.
[tex]\large {\boxed {^nC_r = \frac{n!}{r! (n - r)!} } }[/tex]
Let us tackle the problem.
There are four nickels and seven dimes in your pocket.
The probability of selecting a dime P(D) is :
[tex]P(D) = \frac{\text{Number of dimes}}{\text {Total Number of coins}}[/tex]
[tex]P(D) = \frac{7}{7+4}[/tex]
[tex]P(D) = \boxed {\frac{7}{11}}[/tex]
One of the nickels and one of the dimes are Canadian.
The probability of selecting Canadian Currency P(C) is :
[tex]P(C) = \frac{\text{Number of Canadian Currency}}{\text {Total Number of coins}}[/tex]
[tex]P(C) = \frac{1 + 1}{7+4}[/tex]
[tex]P(C) = \boxed {\frac{2}{11} }[/tex]
The probability of selecting a dime and Canadian Currency P(D ∩ C) is :
[tex]P(C) = \frac{\text{Number of Canadian dimes}}{\text {Total Number of coins}}[/tex]
[tex]P(C) = \frac{1}{7+4}[/tex]
[tex]P(C) = \boxed {\frac{1}{11}}[/tex]
The probability of selecting a dime or Canadian Currency P(D ∪ C) is :
[tex]P(D \cup C) = P(D) + P(C) - P(D \cap C)[/tex]
[tex]P(D \cup C) = \frac{7}{11} + \frac{2}{11} - \frac{1}{11}[/tex]
[tex]P(D \cup C) = \boxed {\frac{8}{11}}[/tex]
Learn more
- Different Birthdays : https://brainly.com/question/7567074
- Dependent or Independent Events : https://brainly.com/question/12029535
- Mutually exclusive : https://brainly.com/question/3464581
Answer details
Grade: High School
Subject: Mathematics
Chapter: Probability
Keywords: Probability , Sample , Space , Six , Dice , Die , Binomial , Distribution , Mean , Variance , Standard Deviation
