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Umnica [9.8K]
2 years ago
10

A pet store has 11 ​puppies, including 4 ​poodles, 5 ​terriers, and 2 retrievers. If Rebecka and​ Aaron, in that​ order, each se

lect one puppy at random with replacement​ (they may both select the same​ one), find the probability that Rebecka selects a terrier and Aaron selects a retriever.
Mathematics
1 answer:
Nataliya [291]2 years ago
8 0

The probability that Rebecka selects a terrier and Aaron selects a retriever is 10/121

<h3>How to determine the probability?</h3>

The distribution of the puppies is given as:

4 ​poodles, 5 ​terriers, and 2 retrievers

The probability of selecting a terrier is;

P(terrier) = 5/11

The probability of selecting a retriever is;

P(retriever) = 2/11

The required probability is

P = 5/11 * 2/11

Evaluate

P = 10/121

Hence, the probability that Rebecka selects a terrier and Aaron selects a retriever is 10/121

Read more about probability at:

brainly.com/question/11234923

#SPJ1

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Also, we are told that 300 students did not wear school colours green and white.

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percent of the school population who wore their school colors is; (1200/1500) × 100% = 80%

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a) S={(head,1),(head,2),(head,3),(head,4),(head,5),(head,6),(tails,1),(tails,2),(tails,3),(tails,4),(tails,5),(tails,6)}

b) S={top,bottom,side}

c) S={(Ace of diamonds),(Two of diamonds),(Three of diamonds),(Four of diamonds),(Five of diamonds),(Six of diamonds),(Seven of diamonds), (Eight of diamonds),(Nine of diamonds),(Ten of diamonds), (Jack of diamonds), (Queen of diamonds), (King of diamonds), (Ace of hearts),(Two of hearts),(Three of hearts),(Four of hearts),(Five of hearts),(Six of hearts),(Seven of hearts), (Eight of hearts), (Nine of hearts),(Ten of hearts), (Jack of hearts), (Queen of hearts),(King of hearts),(Ace of clubs),(Two of clubs),(Three of clubs),(Four of clubs),(Five of clubs),(Six of clubs),(Seven of clubs), (Eight of clubs), (Nine of clubs),(Ten of clubs), (Jack of clubs),(Queen of clubs),(King of clubs),(Ace of spades),(Two of spades),(Three of spades),(Four of spades),(Five of spades),(Six of spades),(Seven of spades),(Eight of spades), (Nine of spades),(Ten of spades),(Jack of spades),(Queen of spades),(King of spades)}

Step-by-step explanation:

The sample space is all the possible outcomes from each experiment:

a) The sample has the pairs of the result of the coin (head or tails) and the dice (a number from 1 to 6):

S={(head,1),(head,2),(head,3),(head,4),(head,5),(head,6),(tails,1),(tails,2),(tails,3),(tails,4),(tails,5),(tails,6)}

b) A paper cup can land in three different ways, on its top, bottom or  side:

S={top,bottom,side}

c) The sample space is all the cards in the deck:

S={(Ace of diamonds),(Two of diamonds),(Three of diamonds),(Four of diamonds),(Five of diamonds),(Six of diamonds),(Seven of diamonds), (Eight of diamonds),(Nine of diamonds),(Ten of diamonds), (Jack of diamonds), (Queen of diamonds), (King of diamonds), (Ace of hearts),(Two of hearts),(Three of hearts),(Four of hearts),(Five of hearts),(Six of hearts),(Seven of hearts), (Eight of hearts), (Nine of hearts),(Ten of hearts), (Jack of hearts), (Queen of hearts),(King of hearts),(Ace of clubs),(Two of clubs),(Three of clubs),(Four of clubs),(Five of clubs),(Six of clubs),(Seven of clubs), (Eight of clubs), (Nine of clubs),(Ten of clubs), (Jack of clubs),(Queen of clubs),(King of clubs),(Ace of spades),(Two of spades),(Three of spades),(Four of spades),(Five of spades),(Six of spades),(Seven of spades),(Eight of spades), (Nine of spades),(Ten of spades),(Jack of spades),(Queen of spades),(King of spades)}

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