Question

diane had 4 cans of green beans, 3 cans of corn, 6 cans of green peas, 3 cans of carrots and 4 cans of baked beans in her pantry. when her granddaughter came to visit, she tore the labels off every can in the pantry! diane chose one can at random from the unlabeled cans for each night’s meal the next week. what is the probability of diane choosing a can of corn on monday, serving it, and then choosing a can of baked beans on tuesday? a \\(\frac{3}{95}\\) b \\(\frac{9}{190}\\) c \\(\frac{3}{100}\\) d \\(\frac{3}{50}\\)

Explanation:

Step1: Calculate total cans initially

First, find the total number of cans Diane has. The cans are: 4 (green beans) + 3 (corn) + 6 (green peas) + 3 (carrots) + 4 (baked beans). So, \(4 + 3 + 6 + 3 + 4 = 20\) cans.

Step2: Probability of choosing corn on Monday

The number of corn cans is 3. So the probability of choosing corn on Monday is \(\frac{3}{20}\) (since after choosing corn, we don't replace it, so the total number of cans for Tuesday will be 19, and the number of baked beans is 4).

Step3: Probability of choosing baked beans on Tuesday

After choosing a can of corn on Monday, the total number of cans left is \(20 - 1 = 19\). The number of baked beans cans is 4. Wait, no, wait: Wait, the problem is choosing corn on Monday and baked beans on Tuesday. Wait, initial counts: green beans (4), corn (3), green peas (6), carrots (3), baked beans (4). So total initial: 4+3+6+3+4=20.

Probability of corn on Monday: number of corn is 3, so \(P(\text{corn}) = \frac{3}{20}\). Then, after removing one corn can, total cans left: 19. Number of baked beans is 4? Wait, no, the baked beans are 4? Wait, the problem says "4 cans of green beans, 3 cans of corn, 6 cans of green peas, 3 cans of carrots and 4 cans of baked beans". So yes, baked beans are 4. Wait, but the question is probability of choosing corn on Monday and baked beans on Tuesday. So first, probability of corn: 3/20. Then, after taking out a corn can, total cans are 19, and baked beans are 4. So probability of baked beans after corn is 4/19? Wait, no, wait the answer options are 3/95, 9/190, etc. Wait, maybe I misread the problem. Wait, let's re-read: "Diane had 4 cans of green beans, 3 cans of corn, 6 cans of green peas, 3 cans of carrots and 4 cans of baked beans in her pantry. When her granddaughter came to visit, she tore the labels off every can in the pantry! Diane chose one can at random from the unlabeled cans for each night’s meal and then choosing a can of baked beans on Tuesday?" Wait, no, the question is: "What is the probability of Diane choosing a can of corn on Monday, serving it, and then choosing a can of baked beans on Tuesday?"

Ah, right, so first event: choosing corn on Monday. Number of corn cans: 3. Total cans: 4+3+6+3+4=20. So \(P(\text{corn}) = \frac{3}{20}\).

Second event: after serving the corn can, the total number of cans left is 20 - 1 = 19. Number of baked beans cans: 4. So \(P(\text{baked beans after corn}) = \frac{4}{19}\)? Wait, but the answer options are 3/95, 9/190, 3/100, 3/50. Wait, maybe I made a mistake. Wait, maybe the baked beans are 3? Wait, no, the problem says 4 cans of baked beans? Wait, let's check the problem again: "4 cans of green beans, 3 cans of corn, 6 cans of green peas, 3 cans of carrots and 4 cans of baked beans". So baked beans: 4. Corn: 3.

Wait, but 3/20 4/19 = 12/380 = 3/95. Oh! Wait, 34=12, 20*19=380. Simplify 12/380: divide numerator and denominator by 4: 3/95. Which is option A.

Wait, let's recalculate:

Total cans: 4 (green beans) + 3 (corn) + 6 (green peas) + 3 (carrots) + 4 (baked beans) = 4+3=7, 7+6=13, 13+3=16, 16+4=20. Correct.

Probability of corn on Monday: 3/20.

After removing one corn can, total cans: 19.

Number of baked beans: 4. So probability of baked beans on Tuesday: 4/19.

Multiply the two probabilities: (3/20) (4/19) = (34)/(20*19) = 12/380 = 3/95 (divide numerator and denominator by 4: 12÷4=3, 380÷4=95).

Answer:

A. \(\frac{3}{95}\)