Question 1158646
<i><b>a. One red, one black and one red marble (in that order)</i></b>:
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{{{(3/10) * (7/9) * (2/8)}}} = {{{42/720}}} = {{{7/120}}}
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<i><b>b. Two black marbles:</i></b>
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{{{(7C2 * 3C1)/(10C3)}}} = {{{((7!/(2!*5!)) * (3!/(1!*2!)))/(10!/(3!*7!))}}} = {{{(21 * 3)/120}}} = {{{63/120}}} = {{{21/40}}}
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<i><b>c. At least two black marbles:</i></b>
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{{{((7C2 * 3C1) + 7C3)/(10C3)}}} = {{{((7!/(2!*5!)) * (3!/(1!*2!)) + (7!/(3!*4!)))/(10!/(3!*7!))}}} = {{{((21 * 3) + 35)/120}}} = {{{(63 + 35)/120}}} = {{{98/120}}} = {{{49/60}}}
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<i><b>d. At most two black marbles:</i></b>
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{{{1 - (7C3/(10C3))}}} = {{{1 - ((7!/(3!*4!)))/((10!/(3!*7!)))}}} = {{{1 - (35/120)}}} = {{{85/120}}} = {{{17/24}}}