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Answer :
Sure! Let's go through the solution step by step for the given problem:
a) List the sample space.
When we select two cards with replacement from the box, each card can appear in both positions. The cards are marked with a question mark (Q), a pear (P), and a triangle (T). Here’s how you list all the possible outcomes:
1. First card Q, Second card Q - (Q, Q)
2. First card Q, Second card P - (Q, P)
3. First card Q, Second card T - (Q, T)
4. First card P, Second card Q - (P, Q)
5. First card P, Second card P - (P, P)
6. First card P, Second card T - (P, T)
7. First card T, Second card Q - (T, Q)
8. First card T, Second card P - (T, P)
9. First card T, Second card T - (T, T)
Therefore, the sample space is:
[('Q', 'Q'), ('Q', 'P'), ('Q', 'T'), ('P', 'Q'), ('P', 'P'), ('P', 'T'), ('T', 'Q'), ('T', 'P'), ('T', 'T')].
b) Choose the correct answer for the sample space:
The correct answer that matches the above list is option B: QQ, QP, QT, PQ, PP, PT, TQ, TP, TT.
c) Determine the probability that two triangles are selected.
From the sample space, we see that there is only one outcome where both cards are triangles: (T, T).
The probability of this happening is calculated by dividing the number of favorable outcomes by the total number of outcomes in the sample space:
- Number of favorable outcomes (T, T) = 1
- Total number of outcomes = 9
So, the probability is [tex]\( \frac{1}{9} \)[/tex].
d) Determine the probability that a card containing a pear and then a card containing a triangle are selected.
From the sample space, the outcome where a pear is followed by a triangle is (P, T).
- Number of favorable outcomes (P, T) = 1
- Total number of outcomes = 9
Thus, the probability is also [tex]\( \frac{1}{9} \)[/tex].
By breaking down the problem in this way, it’s clear how the probabilities are derived from the sample space of two-card selections with replacement.
a) List the sample space.
When we select two cards with replacement from the box, each card can appear in both positions. The cards are marked with a question mark (Q), a pear (P), and a triangle (T). Here’s how you list all the possible outcomes:
1. First card Q, Second card Q - (Q, Q)
2. First card Q, Second card P - (Q, P)
3. First card Q, Second card T - (Q, T)
4. First card P, Second card Q - (P, Q)
5. First card P, Second card P - (P, P)
6. First card P, Second card T - (P, T)
7. First card T, Second card Q - (T, Q)
8. First card T, Second card P - (T, P)
9. First card T, Second card T - (T, T)
Therefore, the sample space is:
[('Q', 'Q'), ('Q', 'P'), ('Q', 'T'), ('P', 'Q'), ('P', 'P'), ('P', 'T'), ('T', 'Q'), ('T', 'P'), ('T', 'T')].
b) Choose the correct answer for the sample space:
The correct answer that matches the above list is option B: QQ, QP, QT, PQ, PP, PT, TQ, TP, TT.
c) Determine the probability that two triangles are selected.
From the sample space, we see that there is only one outcome where both cards are triangles: (T, T).
The probability of this happening is calculated by dividing the number of favorable outcomes by the total number of outcomes in the sample space:
- Number of favorable outcomes (T, T) = 1
- Total number of outcomes = 9
So, the probability is [tex]\( \frac{1}{9} \)[/tex].
d) Determine the probability that a card containing a pear and then a card containing a triangle are selected.
From the sample space, the outcome where a pear is followed by a triangle is (P, T).
- Number of favorable outcomes (P, T) = 1
- Total number of outcomes = 9
Thus, the probability is also [tex]\( \frac{1}{9} \)[/tex].
By breaking down the problem in this way, it’s clear how the probabilities are derived from the sample space of two-card selections with replacement.
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