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Using rule based probability, A box contains 10 good lemons and 7 bad lemons. Bob is going to randomly draw out 5 lemons. Find the

probability that he will get at least one good lemon.

Answer is something over 6,188

User Spoonraker
by
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1 Answer

3 votes

Answer: 6167/6188

This reduces to 881/884

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Step-by-step explanation:

There are 10 good lemons and 7 bad lemons. That's 10+7 = 17 total.

  • 7/17 = probability of getting a bad lemon as 1st selection.
  • 6/16 = probability of getting a bad lemon as 2nd selection.
  • 5/15 = probability of getting a bad lemon as 3rd selection.
  • 4/14 = probability of getting a bad lemon as 4th selection.
  • 3/13 = probability of getting a bad lemon as 5th selection.

Notice the numerators count down (7,6,5,4,3) and so do the denominators (17,16,15,14,13). This is because each lemon selected is not put back, aka no replacement.

Multiply out those fractions

(7/17)*(6/16)*(5/15)*(4/14)*(3/13)

(7*6*5*4*3)/(17*16*15*14*13)

2520/742560

3/884

The probability of getting 5 bad lemons in a row is exactly 3/884

Subtract that from 1 to determine the probability of getting at least one good lemon. This works because:

P(5 bad lemons) + P(at least one good lemon) = 1

P(at least one good lemon) = 1 - P(5 bad lemons)

So,

P(at least one good lemon) = 1 - P(5 bad lemons)

P(at least one good lemon) = 1 - (3/884)

P(at least one good lemon) = (884/884) - (3/884)

P(at least one good lemon) = (884-3)/884

P(at least one good lemon) = 881/884

That's the fraction when we reduce fully.

But we can multiply top and bottom by 7 to make the denominator 6188

881*7 = 6167

884*7 = 6188

This means that 881/884 = 6167/6188

I'm not sure why your teacher has made the denominator 6188.

User Aequitas
by
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