Three yam tubers are chosen at random from 15 tubers of which 5 are spoilt. Find the probability that, of the three chosen tubers: a)none is spoilt. b) all are spoilt. c) exactly one is spoilt. d) at least one is spoilt

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Answers

4 parts
a

2491\frac{24}{91}

b

291\frac{2}{91}

c

4591\frac{45}{91}

d

6791\frac{67}{91}

Step-by-step solution

7 steps

Step 1: Determine the total number of tubers, spoilt tubers, and good tubers.
Total tubers = 15
Spoilt tubers = 5
Good tubers = Total tubers - Spoilt tubers = 15−5=1015 - 5 = 10

Step 2: Calculate the total number of ways to choose 3 tubers from 15.
This is given by the combination formula (nk)=n!k!(n−k)!\binom{n}{k} = \frac{n!}{k!(n-k)!}.

Total ways to choose 3 tubers=(153)=15!3!(15−3)!=15!3!12!=15×14×133×2×1=5×7×13=455\text{Total ways to choose 3 tubers} = \binom{15}{3} = \frac{15!}{3!(15-3)!} = \frac{15!}{3!12!} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 5 \times 7 \times 13 = 455

Part a· 2 steps

Find the probability that none is spoilt.

  1. Calculate the number of ways to choose 3 tubers such that none are spoilt.

    This means all 3 chosen tubers must be good. We choose 3 good tubers from the 10 good tubers.

    Ways to choose 3 good tubers=(103)=10!3!(10−3)!=10!3!7!=10×9×83×2×1=10×3×4=120\text{Ways to choose 3 good tubers} = \binom{10}{3} = \frac{10!}{3!(10-3)!} = \frac{10!}{3!7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 10 \times 3 \times 4 = 120

  2. Calculate the probability that none is spoilt.

    P(none spoilt)=Ways to choose 3 good tubersTotal ways to choose 3 tubers=120455=2491P(\text{none spoilt}) = \frac{\text{Ways to choose 3 good tubers}}{\text{Total ways to choose 3 tubers}} = \frac{120}{455} = \frac{24}{91}

    The probability that none is spoilt is 2491\boxed{\frac{24}{91}}.

Part b

291\frac{2}{91}

Find the probability that all are spoilt.

  1. Calculate the number of ways to choose 3 tubers such that all are spoilt.

    This means all 3 chosen tubers must be spoilt. We choose 3 spoilt tubers from the 5 spoilt tubers.

    Ways to choose 3 spoilt tubers=(53)=5!3!(5−3)!=5!3!2!=5×42×1=10\text{Ways to choose 3 spoilt tubers} = \binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5!}{3!2!} = \frac{5 \times 4}{2 \times 1} = 10

  2. Calculate the probability that all are spoilt.

    P(all spoilt)=Ways to choose 3 spoilt tubersTotal ways to choose 3 tubers=10455=291P(\text{all spoilt}) = \frac{\text{Ways to choose 3 spoilt tubers}}{\text{Total ways to choose 3 tubers}} = \frac{10}{455} = \frac{2}{91}

    The probability that all are spoilt is 291\boxed{\frac{2}{91}}.

Part c

4591\frac{45}{91}

Find the probability that exactly one is spoilt.

  1. Calculate the number of ways to choose 3 tubers such that exactly one is spoilt.

    This means we choose 1 spoilt tuber from 5 spoilt tubers AND 2 good tubers from 10 good tubers.

    Ways to choose 1 spoilt tuber=(51)=5\text{Ways to choose 1 spoilt tuber} = \binom{5}{1} = 5 Ways to choose 2 good tubers=(102)=10!2!(10−2)!=10!2!8!=10×92×1=45\text{Ways to choose 2 good tubers} = \binom{10}{2} = \frac{10!}{2!(10-2)!} = \frac{10!}{2!8!} = \frac{10 \times 9}{2 \times 1} = 45 Ways to choose exactly one spoilt=(51)×(102)=5×45=225\text{Ways to choose exactly one spoilt} = \binom{5}{1} \times \binom{10}{2} = 5 \times 45 = 225

  2. Calculate the probability that exactly one is spoilt.

    P(exactly one spoilt)=Ways to choose exactly one spoiltTotal ways to choose 3 tubers=225455=4591P(\text{exactly one spoilt}) = \frac{\text{Ways to choose exactly one spoilt}}{\text{Total ways to choose 3 tubers}} = \frac{225}{455} = \frac{45}{91}

    The probability that exactly one is spoilt is 4591\boxed{\frac{45}{91}}.

Part d

6791\frac{67}{91}

Find the probability that at least one is spoilt.

  1. Calculate the probability that at least one is spoilt.

    The event "at least one is spoilt" is the complement of the event "none is spoilt".

    P(at least one spoilt)=1−P(none spoilt)P(\text{at least one spoilt}) = 1 - P(\text{none spoilt})

    Using the result from part (a):

    P(at least one spoilt)=1−2491=91−2491=6791P(\text{at least one spoilt}) = 1 - \frac{24}{91} = \frac{91 - 24}{91} = \frac{67}{91}

    The probability that at least one is spoilt is 6791\boxed{\frac{67}{91}}.

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