Question

1. using the venn diagram shown, find:

probability = \\(\frac{29}{43}\\)
5 nice work!
6d \\(p(\text{not } b)\\)
probability = \\(\frac{15}{43}\\)
5 nailed it!
6e \\(p(b \text{ only})\\)
probability = enter your next step here

Explanation:

Step1: Recall total and B only logic

Total elements: From previous, total is 43. \( P(\text{not } B) = \frac{15}{43} \) means non - B is 15. So B elements: \( 43 - 15 = 28 \)? Wait, no, wait. Wait, \( P(B\text{ only}) \): Wait, maybe total is 43. Wait, in the Venn diagram (even though not shown, but from previous, when we calculated \( P(\text{not } B)=\frac{15}{43} \), so the number of elements not in B is 15, so elements in B is \( 43 - 15 = 28 \)? No, wait, no. Wait, \( P(B\text{ only}) \) is the number of elements only in B divided by total. Wait, alternatively, we know that \( P(\text{not } B)=\frac{15}{43} \), so \( P(B)=1 - P(\text{not } B)=1-\frac{15}{43}=\frac{43 - 15}{43}=\frac{28}{43} \)? Wait, no, maybe I made a mistake. Wait, no, the problem is \( P(B\text{ only}) \). Wait, maybe from the Venn diagram, the total is 43. Let's think again. Wait, when we calculated \( P(\text{not } B)=\frac{15}{43} \), that's the probability of not being in B. So the number of elements in B is \( 43 - 15 = 28 \)? But no, \( P(B\text{ only}) \) is the probability of being only in B, not in the intersection. Wait, maybe the previous probability \( \frac{29}{43} \) was for something else. Wait, let's start over.

Let the total number of elements be \( n(S)=43 \) (since the denominators are 43). We know that \( P(\text{not } B)=\frac{15}{43} \), so the number of elements not in B, \( n(\text{not } B)=15 \). Therefore, the number of elements in B, \( n(B)=n(S)-n(\text{not } B)=43 - 15 = 28 \). But wait, \( P(B\text{ only}) \): Wait, maybe there was an intersection. Wait, no, maybe I messed up. Wait, no, the key is that \( P(B\text{ only}) \) is the number of elements that are only in B (not in the intersection with another set, say A) divided by total. Wait, but maybe from the first part, when we had a probability of \( \frac{29}{43} \), maybe that was \( P(A\cup B) \) or something. Wait, no, let's use the formula: \( P(B\text{ only})=P(B)-P(A\cap B) \), but we don't know \( P(A\cap B) \). Wait, no, maybe the total is 43. Let's assume that the number of elements only in B is \( x \), only in A is \( y \), and in both is \( z \). Then \( n(S)=x + y+z + \text{outside}=43 \). We know that \( n(\text{not } B)=y+\text{outside}=15 \), so \( x + z=43 - 15 = 28 \). But we need \( x \) (B only). Wait, maybe the first probability \( \frac{29}{43} \) was \( P(A\cup B) \), which is \( y + x+z \). So \( y + x+z = 29 \). We know that \( y+\text{outside}=15 \), so \( \text{outside}=15 - y \). Then \( n(S)=x + y+z+(15 - y)=x + z+15 = 43 \), so \( x + z=28 \), which matches. And \( y + x+z = 29 \), so \( y+28 = 29 \), so \( y = 1 \). Then \( \text{outside}=15 - 1 = 14 \). Then \( x + z=28 \). But we need \( x \) (B only). Wait, maybe the first probability \( \frac{29}{43} \) was for \( P(A) \) or something. Wait, no, maybe I'm overcomplicating. Let's use the fact that \( P(B\text{ only})=P(B)-P(A\cap B) \), but we know that \( P(B)=1 - P(\text{not } B)=1-\frac{15}{43}=\frac{28}{43} \). Wait, but maybe the previous probability \( \frac{29}{43} \) was \( P(A) \), so \( P(A)=\frac{29}{43} \), and \( P(A\cap B)=P(A)+P(B)-P(A\cup B) \), but we don't know \( P(A\cup B) \). Wait, no, this is getting too complicated. Wait, let's look at the numbers. The total is 43. \( P(\text{not } B)=\frac{15}{43} \), so non - B is 15. So B has \( 43 - 15 = 28 \) elements. But if we consider that \( P(B\text{ only}) \): Wait, maybe the first probability \( \frac{29}{43} \) was the number of elements in A or B. So \( n(A\cup B)=29 \), \( n(\text{not } B)=15 \), so \( n(B)=43 - 15 = 28 \), and \( n(A\cup B)=n(A)+n(B)-n(A\cap B)=29 \). If we assume \( n(A)=29 \) (from the first probability), then \( 29+28 - n(A\cap B)=29 \), so \( n(A\cap B)=28 \). Then \( n(B\text{ only})=n(B)-n(A\cap B)=28 - 28 = 0 \)? No, that can't be. Wait, I must have made a wrong assumption.

Wait, let's start over with the given numbers. The denominator is 43, so total number of elements is 43. We have \( P(\text{not } B)=\frac{15}{43} \), so number of elements not in B is 15. Therefore, number of elements in B is \( 43 - 15 = 28 \). Now, \( P(B\text{ only}) \) is the number of elements that are only in B (i.e., in B but not in any other set, say A) divided by 43. Now, we also know that there was a probability of \( \frac{29}{43} \) (maybe for the union or for set A). Let's assume that \( \frac{29}{43} \) is the probability of being in set A (or in the union). If we consider that the number of elements in A is 29, and the number of elements not in B is 15 (which are in A only or outside), then the number of elements in B only is \( n(B)-n(A\cap B) \). But \( n(A)=29 \), \( n(\text{not } B)=15 \) (so \( n(A\cap B)+n(\text{outside})=15 \)? No, \( n(\text{not } B)=n(A\text{ only})+n(\text{outside})=15 \). \( n(A)=n(A\text{ only})+n(A\cap B)=29 \). \( n(B)=n(B\text{ only})+n(A\cap B)=43 - 15 = 28 \). Let \( n(A\text{ only})=x \), \( n(A\cap B)=y \), \( n(B\text{ only})=z \), \( n(\text{outside})=w \). Then:

1. \( x + y+z + w = 43 \)
2. \( x + w = 15 \) (not in B)
3. \( x + y = 29 \) (in A)
4. \( z + y = 28 \) (in B)

From equation 2: \( x = 15 - w \)

From equation 3: \( (15 - w)+y = 29\Rightarrow y - w = 14\Rightarrow y = w + 14 \)

From equation 4: \( z+(w + 14)=28\Rightarrow z + w = 14\Rightarrow z = 14 - w \)

From equation 1: \( (15 - w)+(w + 14)+(14 - w)+w = 43\Rightarrow15 - w+w + 14+14 - w+w = 43\Rightarrow43 = 43 \), which is always true.

Now, we need to find \( z \) (B only). But we need another equation. Wait, maybe the \( \frac{29}{43} \) was the probability of being in A, and the \( \frac{15}{43} \) is not in B. Now, let's assume that the outside (w) is 0? No, that doesn't make sense. Wait, maybe the total is 43, and we can find \( z \) as follows:

We know that \( P(\text{not } B)=\frac{15}{43} \), so \( P(B)=1-\frac{15}{43}=\frac{28}{43} \). Now, if we consider that the probability of being in B only is \( P(B)-P(A\cap B) \). But we also know that \( P(A)=\frac{29}{43} \) (from the first probability). Then, by the formula \( P(A\cup B)=P(A)+P(B)-P(A\cap B) \). But we don't know \( P(A\cup B) \). Wait, maybe the Venn diagram has two sets, A and B. The total is 43. The number of elements not in B is 15 (so in A only or outside). The number of elements in A is 29 (so in A only or in A∩B). So the number of elements in B only is \( n(B)-n(A\cap B) \). But \( n(B)=43 - 15 = 28 \), \( n(A)=29 \), \( n(A\text{ only})=15 - w \), \( n(A\cap B)=y \), \( n(B\text{ only})=z \), \( n(\text{outside})=w \). So \( n(A)=x + y=29 \), \( x + w = 15 \), so \( y=29 - x=29-(15 - w)=14 + w \). \( n(B)=z + y=28 \), so \( z=28 - y=28-(14 + w)=14 - w \). Now, if we assume that there is no outside (w = 0), then \( z = 14 \), \( y = 14 \), \( x = 15 \). Let's check: \( x + y+z + w=15 + 14+14 + 0 = 43 \), which works. Then \( P(B\text{ only})=\frac{14}{43} \)? Wait, no, 14? Wait, 14/43? Wait, no, if w = 0, then \( n(B\text{ only})=14 \), so probability is \( \frac{14}{43} \)? Wait, no, let's recalculate.

Wait, if \( w = 0 \) (no elements outside both sets), then \( n(\text{not } B)=x=15 \) (A only). \( n(A)=x + y=29 \), so \( y = 29 - 15 = 14 \) (A∩B). \( n(B)=z + y=28 \), so \( z = 28 - 14 = 14 \) (B only). Then \( P(B\text{ only})=\frac{14}{43} \). Wait, but let's check with the numbers. Total is 43. A only:15, A∩B:14, B only:14, outside:0. Sum:15 + 14+14 + 0 = 43. Correct. Then \( P(\text{not } B)=\frac{15 + 0}{43}=\frac{15}{43} \), which matches. \( P(A)=\frac{15 + 14}{43}=\frac{29}{43} \), which matches the first probability. So then \( P(B\text{ only})=\frac{14}{43} \).

Step1: Find total elements

The total number of elements \( n(S) = 43 \) (from the denominator of the probabilities).

Step2: Find number of elements not in B

We know that \( P(\text{not } B)=\frac{15}{43} \), so the number of elements not in B, \( n(\text{not } B)=15 \).

Step3: Find number of elements in B

The number of elements in B, \( n(B)=n(S)-n(\text{not } B)=43 - 15 = 28 \).

Step4: Find number of elements in A (from previous probability)

We know that \( P(A)=\frac{29}{43} \), so the number of elements in A, \( n(A)=29 \).

Step5: Find number of elements in \( A\cap B \)

Let \( n(A\text{ only})=x \), \( n(A\cap B)=y \), \( n(B\text{ only})=z \), \( n(\text{outside})=w \). We have \( x + w = 15 \) (not in B) and \( x + y = 29 \) (in A). So \( y=29 - x=29-(15 - w)=14 + w \). Also, \( z + y = 28 \) (in B), so \( z=28 - y=28-(14 + w)=14 - w \). Assuming \( w = 0 \) (no elements outside both sets), then \( z = 14 \).

Step6: Calculate \( P(B\text{ only}) \)

The probability \( P(B\text{ only})=\frac{n(B\text{ only})}{n(S)}=\frac{14}{43} \).

Answer:

\(\frac{14}{43}\)