Question

for classroom awards, ms. blake has coupons in a bag that students can draw at random. in the bag, she has 6 coupons that say “do half the assignment” and 4 coupons that say “no homework”. what is the probability of sally drawing a “no homework” coupon, keeping it, and then charlie drawing a “no homework” coupon? a \\(\frac{4}{25}\\) b \\(\frac{2}{15}\\) c \\(\frac{6}{25}\\) d \\(\frac{1}{3}\\)

Explanation:

Step1: Determine total coupons initially

There are 6 "Do Half the Assignment" coupons and 4 "No Homework" coupons. So total initial coupons: \(6 + 4 = 10\).

Step2: Probability Sally draws "No Homework"

Sally draws first. Number of "No Homework" coupons is 4, total is 10. So probability for Sally: \(\frac{4}{10}=\frac{2}{5}\).

Step3: Update total and "No Homework" coupons after Sally

After Sally draws a "No Homework" coupon, total coupons left: \(10 - 1 = 9\), "No Homework" coupons left: \(4 - 1 = 3\).

Step4: Probability Charlie draws "No Homework"

Now Charlie draws. Probability: \(\frac{3}{9}=\frac{1}{3}\).

Step5: Multiply probabilities for both events

We need both Sally and Charlie to draw "No Homework" coupons. So multiply their probabilities: \(\frac{2}{5}\times\frac{1}{3}=\frac{2}{15}\). Wait, no—wait, the question is "Sally drawing a 'No Homework' coupon, keeping it, and then Charlie drawing a 'No Homework' coupon?" Wait, no, re-reading: Wait, the original problem: Wait, maybe I misread. Wait, the problem says: "What is the probability of Sally drawing a 'No Homework' coupon, keeping it, and then Charlie drawing a 'No Homework' coupon?" Wait, no, let's check again. Wait, the problem: "For classroom awards, Ms. Blake has coupons in a bag that students can draw at random. In the bag, she has 6 coupons that say 'Do Half the Assignment' and 4 coupons that say 'No Homework'. What is the probability of Sally drawing a 'No Homework' coupon, keeping it, and then Charlie drawing a 'No Homework' coupon?" Wait, so first Sally draws, then Charlie. So:

Total coupons: 6 + 4 = 10.

Sally's probability: 4/10 = 2/5.

After Sally takes one "No Homework" coupon, total left: 9, "No Homework" left: 3.

Charlie's probability: 3/9 = 1/3.

Now, the probability of both events (Sally then Charlie) is (2/5)(1/3) = 2/15? Wait, but the options: Option B is 2/15. Wait, but let's check the options again. Wait, the options are A. 4/25, B. 2/15, C. 6/25, D. 1/3. Wait, maybe I made a mistake. Wait, no—wait, maybe the problem is "Sally drawing a 'No Homework' coupon" and then "Charlie drawing a 'No Homework' coupon"—so the combined probability is (4/10)(3/9) = (2/5)*(1/3) = 2/15, which is option B. Wait, but let's re-express:

Wait, total coupons: 6 (Do Half) + 4 (No Homework) = 10.

Sally draws "No Homework": 4/10 = 2/5.

Then, total coupons left: 9, "No Homework" left: 3.

Charlie draws "No Homework": 3/9 = 1/3.

Multiply: (2/5)*(1/3) = 2/15. So the answer is B. 2/15.

Wait, but maybe the problem was misread. Wait, the original problem: "What is the probability of Sally drawing a 'No Homework' coupon, keeping it, and then Charlie drawing a 'No Homework' coupon?" So yes, that's the calculation. So the correct option is B. 2/15.

Answer:

B. \(\frac{2}{15}\)