A spinner has numbers 1-11 marked equally on the face. If the spinner is spun 3 times, what is the probability of getting an even and an odd and a 4?
A) $\frac { 30 } { 1,331 }$
B) $\frac { 1 } { 121 }$
C) $\frac { 15 } { 671 }$
D) $\frac { 1 } { 1,331 }$
E) $\frac { 1 } { 44 }$

Question

Quizplus · Accepted Answer

The total number of outcomes when spinning the spinner 3 times is $11^3 = 1,331$. To get an even, an odd, and a 4 in any order, we have several scenarios. There are 5 even numbers (2, 4, 6, 8, 10) and 6 odd numbers (1, 3, 5, 7, 9, 11) on the spinner. For the combination including a 4, we have:1. 4 (even), odd, even: $1 \times 6 \times 4 = 24$ ways (since 4 is fixed, 6 options for odd, and 4 remaining evens)2. 4 (even), even, odd: $1 \times 4 \times 6 = 24$ ways (4 options for the other even, and 6 for the odd)3. Even, 4 (even), odd: $4 \times 1 \times 6 = 24$ ways4. Even, odd, 4 (even): $4 \times 6 \times 1 = 24$ ways5. Odd, 4 (even), even: $6 \times 1 \times 4 = 24$ ways6. Odd, even, 4 (even): $6 \times 4 \times 1 = 24$ waysAdding these gives $24 \times 6 = 144$ ways. However, we've double-counted the scenarios where the other even number is also a 4. There are 3 ways this could happen (4, 4, odd), (4, odd, 4), and (odd, 4, 4), each with 6 possibilities for the odd number, so 18 in total. Subtracting these gives $144 - 18 = 126$ correct ways.However, the initial calculation did not account for the specific constraint of including exactly one 4, leading to an overestimation. The correct approach is to consider each unique placement of the 4, the even number (excluding 4), and the odd number:1. Placing the 4, there are 3 choices for its position.2. For the even number (excluding 4), there are 4 choices (2, 6, 8, 10).3. For the odd number, there are 6 choices (1, 3, 5, 7, 9, 11).This gives a total of $3 \times 4 \times 6 = 72$ correct arrangements. However, this calculation still doesn't align with the provided options, indicating a misunderstanding in the approach to calculating the specific combinations that include one 4, one even, and one odd number in any order.Given the options provided and the need for clarification, the correct calculation should focus on the probability of achieving one even number, one odd number, and the number 4 in any order across three spins, considering all possible permutations of these outcomes.The correct answer, based on the options provided and understanding the probability of specific outcomes in a scenario with multiple spins of a spinner with numbers 1-11, would be recalculated considering the total possible outcomes and the specific ways to achieve the desired combination of one even, one odd, and one 4. The initial explanation attempted to break down the possibilities but did not accurately calculate the probability based on the conditions and options given.

A Spinner Has Numbers 1-11 Marked Equally on the Face 301,331\frac { 30 } { 1,331 }1,33130​

A Spinner Has Numbers 1-11 Marked Equally on the Face $\frac { 30 } { 1,331 }$