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Math  /  Data & Statistics

QuestionGiven that Kieran saw exactly one of these types of animal, what is the probability that he saw a lion? Give your answer as a fraction in its simplest form.
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Studdy Solution

STEP 1

1. Kieran saw exactly one type of animal, either a lion or a rhino, but not both.
2. The probability tree diagram provides the probabilities for seeing a lion or a rhino.

STEP 2

1. Identify the relevant probabilities from the tree diagram.
2. Calculate the probability of seeing exactly one type of animal.
3. Calculate the probability of seeing a lion given that exactly one type of animal was seen.

STEP 3

Identify the probabilities from the tree diagram: - Probability of seeing a lion: P(Lion)=25 P(\text{Lion}) = \frac{2}{5} - Probability of not seeing a lion: P(Not Lion)=35 P(\text{Not Lion}) = \frac{3}{5} - Probability of seeing a rhino: P(Rhino)=56 P(\text{Rhino}) = \frac{5}{6} - Probability of not seeing a rhino: P(Not Rhino)=16 P(\text{Not Rhino}) = \frac{1}{6}

STEP 4

Calculate the probability of seeing exactly one type of animal: - Probability of seeing only a lion (and not a rhino): P(Lion and Not Rhino)=P(Lion)×P(Not Rhino)=25×16=230=115 P(\text{Lion and Not Rhino}) = P(\text{Lion}) \times P(\text{Not Rhino}) = \frac{2}{5} \times \frac{1}{6} = \frac{2}{30} = \frac{1}{15} - Probability of seeing only a rhino (and not a lion): P(Rhino and Not Lion)=P(Rhino)×P(Not Lion)=56×35=1530=12 P(\text{Rhino and Not Lion}) = P(\text{Rhino}) \times P(\text{Not Lion}) = \frac{5}{6} \times \frac{3}{5} = \frac{15}{30} = \frac{1}{2}

STEP 5

Calculate the probability that Kieran saw a lion given that exactly one type of animal was seen: - Total probability of seeing exactly one type of animal: P(Exactly One)=P(Lion and Not Rhino)+P(Rhino and Not Lion)=115+12=115+1530=230+1530=1730 P(\text{Exactly One}) = P(\text{Lion and Not Rhino}) + P(\text{Rhino and Not Lion}) = \frac{1}{15} + \frac{1}{2} = \frac{1}{15} + \frac{15}{30} = \frac{2}{30} + \frac{15}{30} = \frac{17}{30} - Probability that Kieran saw a lion given exactly one type of animal: P(Lion∣Exactly One)=P(Lion and Not Rhino)P(Exactly One)=1151730=115×3017=217 P(\text{Lion} \mid \text{Exactly One}) = \frac{P(\text{Lion and Not Rhino})}{P(\text{Exactly One})} = \frac{\frac{1}{15}}{\frac{17}{30}} = \frac{1}{15} \times \frac{30}{17} = \frac{2}{17}
The probability that Kieran saw a lion given that he saw exactly one type of animal is:
217 \boxed{\frac{2}{17}}

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