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attashe74 [19]
3 years ago
10

There are four nickels and seven dimes in your pocket. One of the nickels and one of the dimes are Canadian. The others are US c

urrency. You randomly select a coin from your pocket. It is a dime or is Canadian currency. Find the probability of this occuring.

Mathematics
2 answers:
natima [27]3 years ago
8 0

Answer:

  8/11

Step-by-step explanation:

Only 3 of the nickels are neither dimes nor Canadian. The other 8 of 11 coins are dimes or Canadian. The probability of choosing one of them at random is 8/11.

Rudiy273 years ago
3 0

The probability of selecting a dime or Canadian currency is 8/11

<h3>Further explanation</h3>

The probability of an event is defined as the possibility of an event occurring against sample space.

\large { \boxed {P(A) = \frac{\text{Number of Favorable Outcomes to A}}{\text {Total Number of Outcomes}} } }

<h2>Permutation ( Arrangement )</h2>

Permutation is the number of ways to arrange objects.

\large {\boxed {^nP_r = \frac{n!}{(n - r)!} } }

<h2>Combination ( Selection )</h2>

Combination is the number of ways to select objects.

\large {\boxed {^nC_r = \frac{n!}{r! (n - r)!} } }

Let us tackle the problem.

There are four nickels and seven dimes in your pocket.

<em>The probability of selecting a dime P(D) is :</em>

P(D) = \frac{\text{Number of dimes}}{\text {Total Number of coins}}

P(D) = \frac{7}{7+4}

P(D) = \boxed {\frac{7}{11}}

One of the nickels and one of the dimes are Canadian.

<em>The probability of selecting Canadian Currency P(C) is :</em>

P(C) = \frac{\text{Number of Canadian Currency}}{\text {Total Number of coins}}

P(C) = \frac{1 + 1}{7+4}

P(C) = \boxed {\frac{2}{11} }

<em>The probability of selecting a dime and Canadian Currency P(D ∩ C) is :</em>

P(C) = \frac{\text{Number of Canadian dimes}}{\text {Total Number of coins}}

P(C) = \frac{1}{7+4}

P(C) = \boxed {\frac{1}{11}}

The probability of selecting a dime or Canadian Currency P(D ∪ C) is :

P(D \cup C) = P(D) + P(C) - P(D \cap C)

P(D \cup C) = \frac{7}{11} + \frac{2}{11} - \frac{1}{11}

P(D \cup C) = \boxed {\frac{8}{11}}

<h3>Learn more</h3>
  • Different Birthdays : brainly.com/question/7567074
  • Dependent or Independent Events : brainly.com/question/12029535
  • Mutually exclusive : brainly.com/question/3464581

<h3>Answer details</h3>

Grade: High School

Subject: Mathematics

Chapter: Probability

Keywords: Probability , Sample , Space , Six , Dice , Die , Binomial , Distribution , Mean , Variance , Standard Deviation

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