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Bunuel
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M03-26 25 Jan 2025, 08:34
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miag
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A shop had 10 bread loaves, 7 of which were baguettes. If every loaf has an equal chance of being sold, what is the probability that out of the 6 loaves sold, exactly 4 were baguettes?

A. \(\frac{2}{5}\)
B. \(\frac{3}{5}\)
C. \(\frac{2}{3}\)
D. \(\frac{1}{2}\)
E. \(\frac{4}{7}\)
hi, why does 0.1 probability not matter/doesnt have to be taken into account here?
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The fact that every loaf has an equal chance of being sold doesn’t mean the probability of each loaf being sold is 0.1. It simply means that no specific loaf is favored over another, so the loaves are sold randomly.
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Schools: Anderson '26 (A) Haas '26 (A) Kellogg '26 (A)
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M03-26 25 Jan 2025, 08:50
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Got it, understood now - thank you so much!
Bunuel
miag
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A shop had 10 bread loaves, 7 of which were baguettes. If every loaf has an equal chance of being sold, what is the probability that out of the 6 loaves sold, exactly 4 were baguettes?

A. \(\frac{2}{5}\)
B. \(\frac{3}{5}\)
C. \(\frac{2}{3}\)
D. \(\frac{1}{2}\)
E. \(\frac{4}{7}\)
hi, why does 0.1 probability not matter/doesnt have to be taken into account here?

The fact that every loaf has an equal chance of being sold doesn’t mean the probability of each loaf being sold is 0.1. It simply means that no specific loaf is favored over another, so the loaves are sold randomly.
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M03-26 13 Sep 2025, 10:44
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Hi,

Could you explain why we need to consider non baguettes here?
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Bunuel
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M03-26 13 Sep 2025, 10:48
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MegB07
Hi,

Could you explain why we need to consider non baguettes here?
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Because the question asks for exactly 4 baguettes out of 6 sold.

That means the other 2 loaves must be non-baguettes. To get the total number of favorable cases, you need to count both parts together: choosing 4 from the 7 baguettes and 2 from the 3 non-baguettes. Without including the non-baguette choices, you’d only be counting part of the scenario.
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M03-26 10 Feb 2026, 00:43
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I like the solution - it’s helpful.
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