Question

algebra i b-cr
a bag contains eleven equally sized marbles, which are numbered. two marbles are chosen at random and replaced after each selection.
what is the probability that the first marble chosen is shaded and the second marble chosen is labeled with an odd number?
$\bigcirc \frac{10}{121}$
$\bigcirc \frac{24}{121}$
$\bigcirc \frac{6}{11}$
$\bigcirc \frac{10}{11}$

Explanation:

Step1: Count shaded marbles

Total marbles = 11, shaded marbles = 5 (1,3,4,9, purple ones)
Probability (shaded first) = $\frac{5}{11}$

Step2: Count odd-numbered marbles

Odd-numbered marbles: 1,3,5,7,9,11 (total 6)
Probability (odd second) = $\frac{6}{11}$

Step3: Multiply independent probabilities

Since marbles are replaced, events are independent.
Combined probability = $\frac{5}{11} \times \frac{6}{11} = \frac{30}{121}$
Wait correction: Re-count shaded marbles: Looking at the image, shaded marbles are 1,3,4,9, and one more? Wait no, count again: <point>106 305</point>, <point>226 275</point>, <point>274 327</point>, <point>225 395</point> — that's 4? No, wait the purple ones: 1,3,4,9, and wait <point>105 306</point> (1), <point>229 276</point> (3), <point>277 329</point> (4), <point>226 396</point> (9), and wait <point>102 365</point> is white, <point>154 366</point> white, <point>153 425</point> white, <point>205 328</point> white, <point>258 366</point> white, <point>257 426</point> white. Oh! Shaded marbles are 4? Wait no, the question says "shaded" — the purple circles: 1,3,4,9: 4 marbles. Wait no, wait the first marble shaded: 4 shaded marbles? Then probability shaded first is $\frac{4}{11}$. Then odd numbers: 1,3,5,7,9,11: 6. So $\frac{4}{11} \times \frac{6}{11} = \frac{24}{121}$. Yes, that matches the option. I miscounted shaded marbles earlier. Correct shaded count is 4 (purple circles: 1,3,4,9).

Step1: Correct shaded marble count

Shaded marbles = 4, total =11
$P(\text{shaded first}) = \frac{4}{11}$

Step2: Odd-numbered marble count

Odd marbles =6, total=11
$P(\text{odd second}) = \frac{6}{11}$

Step3: Calculate joint probability

$P = \frac{4}{11} \times \frac{6}{11} = \frac{24}{121}$

Answer:

$\frac{24}{121}$ (Option: $\boldsymbol{\frac{24}{121}}$)