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elixir [45]
4 years ago
15

a survey of 46 college athletes found that 24 played volleyball, while 22 played basketball. if we pick 2 athletes at random wit

hout replacement, what is the probability we get one volleyball and one basketball player
Mathematics
1 answer:
Yuliya22 [10]4 years ago
4 0

Answer:

<u>88/345 which is 0.2551</u>

Step-by-step explanation:

Since we are picking 2 athletes randomly and without replacement.

Probability of 1 volleyball player = 24/46 = 12/23

Probability of basketball player (<u>this</u><u> </u><u>is</u><u> </u><u>aft</u><u>er</u><u> </u><u>1</u><u> </u><u>vol</u><u>ley</u><u>ball</u><u> </u><u>play</u><u>er</u><u> </u><u>has</u><u> </u><u>alr</u><u>eady</u><u> </u><u>be</u><u>en</u><u> </u><u>picked</u><u>)</u><u>.</u><u> </u><u>We</u><u> </u><u>hav</u><u>e</u><u> </u><u>4</u><u>5</u><u> </u><u>players</u><u> </u><u>left</u><u>.</u><u> </u><u>=</u><u> </u><u>2</u><u>2</u><u>/</u><u>4</u><u>5</u>

<u>proba</u><u>bility</u><u> </u><u>of</u><u> </u><u>picki</u><u>ng</u><u> </u><u>one</u><u> </u><u>volle</u><u>yball</u><u> </u><u>pla</u><u>yer</u><u> and</u><u> </u><u>one</u><u> </u><u>bask</u><u>e</u><u>tball</u><u> player</u>

<u>p</u><u>(</u><u>voll</u><u>eyball</u><u> </u><u>pla</u><u>yer</u><u> </u><u>a</u><u>nd</u><u> </u><u>bask</u><u>etball</u><u> </u><u>player</u><u>)</u>

<u>(</u><u>1</u><u>2</u><u>/</u><u>2</u><u>3</u><u>)</u><u> </u><u>x</u><u> </u><u>(</u><u>2</u><u>2</u><u>/</u><u>4</u><u>5</u><u>)</u>

<u>=</u><u> </u><u>8</u><u>8</u><u>/</u><u>3</u><u>4</u><u>5</u>

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