Answer:
1/11
Step-by-step explanation:
<u><em>Given:</em></u>
<em>Bonnie has 4 sharpened and 8 unsharpened pencils in her pencil case. she randomly selects 2 of the pencils from the box without replacement. </em>
<u><em>Question to Answer:</em></u>
<em>what is the probability that both pencils will be sharpened?</em>
<u><em>Solve:</em></u>
<em>Probability is possibility of an event being equal to the ratio of the number of outcomes and the total number of outcomes.</em>
<em>Thus we known that,</em>
<em>Bonnie has 4 sharpened and 8 unsharpened pencils.</em>
<em>Hence, the total numbers of pencils is 12.</em>
<em />
<em>Let, the probability first one is sharpened be P(E₁) and probability second one is sharpened be P(E₂)</em>
<em />
<em>Simplify - P(E₁) = 4/12 = 1/3 and P(E₂) = 3/11</em>
<em />
<em>Therefore we have;</em>
<em> P(E) = P(E₁)×P(E₂)</em>
<em>P(E) = 1/3 × 3/11</em>
<em>P(E) = 3/33</em>
<em>P(E) = 1/11</em>
<em />
<em>As a result, the probability is 1/11 that both pencils will be sharpened.</em>
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<u><em>Kavinsky</em></u>
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