I've been stuck on this question for an hour. whoever gets it gets brainliest answer betty picks 4 random marbles from a bowl containing 3 white, 4 yellow and 5 blue marbles. the probability that exactly 1 of the 4 marbles drawn in blue is {options: 0.0707 0.2135 0.3535 0.6465} the probability that at least 1 of the 4 marbles drawn is white is {options: 0.2181 0.2546 0.5090 0.7455}

Lisa has a great explanation for part B already, so I'll only focus on part A



We have 3 white, 4 yellow and 5 blue. So there are 3+4+5 = 12 marbles total

The probability of picking blue for the first selection is 5/12

The next selection has the probability of 7/11 because there are 7 non-blue marbles (3 white+4 yellow) out of 12-1 = 11 left over

The next selection after that is going to have a probability of 6/10. Simply subtract 1 from the numerator and denominator

Similarly, we go from 6/10 to 5/9

So if we wanted to know the probability of getting blue in slot 1, and non-blue everywhere else, then we multiply out the fractions
(5/12)*(7/11)*(6/10)*(5/9) = 0.08838



We have 4 cases
Case 1) Blue is in slot 1, non-blue everywhere else
Case 2) Blue is in slot 2, non-blue everywhere else
Case 3) Blue is in slot 3, non-blue everywhere else
Case 4) Blue is in slot 4, non-blue everywhere else

when we calculated 0.08838 earlier, we found the probability for case (1) to happen. We want to account for all four cases. Simply multiply 0.08838 by 4 to do this
4*0.08838 = 0.35352 which rounds to 0.3535




So that's why the answer is choice C) 0.3535