To determine the probability of drawing 2 washers, then 3 nuts, and finally 4 bolts from the box without replacement, we need to calculate the probability of each event occurring in sequence and then multiply these probabilities together. This is a problem of sequential probability without replacement.

Understanding the Items in the Box

First, let's identify the total number of items in the box and the count of each type of item:

• Number of washers: 2
• Number of nuts: 3
• Number of bolts: 4
• Total number of items in the box: \(2 + 3 + 4 = 9\)

Probability of Drawing 2 Washers First

We are drawing items one at a time without replacement. The probability of drawing the first washer is the number of washers divided by the total number of items. For the second washer, both the number of washers and the total items decrease by one.

• Probability of drawing the first washer: $$P(\text{1st washer}) = \frac{\text{Number of washers}}{\text{Total items}} = \frac{2}{9}$$
• After drawing one washer, there is 1 washer left and 8 total items.
• Probability of drawing the second washer (given the first was a washer): $$P(\text{2nd washer | 1st washer}) = \frac{\text{Remaining washers}}{\text{Remaining total items}} = \frac{1}{8}$$
• The combined probability of drawing 2 washers first is: $$P(\text{2 washers first}) = P(\text{1st washer}) \times P(\text{2nd washer | 1st washer})$$ $$P(\text{2 washers first}) = \frac{2}{9} \times \frac{1}{8} = \frac{2}{72} = \frac{1}{36}$$

Probability of Drawing 3 Nuts Subsequently

After 2 washers have been drawn, the box now contains 7 items (3 nuts and 4 bolts). We then draw 3 nuts one after another without replacement.

• Number of items remaining after 2 washers are drawn: \(9 - 2 = 7\)
• Number of nuts remaining: 3
• Probability of drawing the first nut: $$P(\text{1st nut}) = \frac{\text{Number of nuts}}{\text{Remaining total items}} = \frac{3}{7}$$
• After drawing one nut, there are 2 nuts left and 6 total items.
• Probability of drawing the second nut (given the first was a nut): $$P(\text{2nd nut | 1st nut}) = \frac{\text{Remaining nuts}}{\text{Remaining total items}} = \frac{2}{6}$$
• After drawing two nuts, there is 1 nut left and 5 total items.
• Probability of drawing the third nut (given the first two were nuts): $$P(\text{3rd nut | 2nd nut}) = \frac{\text{Remaining nuts}}{\text{Remaining total items}} = \frac{1}{5}$$
• The combined probability of drawing 3 nuts subsequently is: $$P(\text{3 nuts subsequently}) = \frac{3}{7} \times \frac{2}{6} \times \frac{1}{5} = \frac{6}{210} = \frac{1}{35}$$

Probability of Drawing 4 Bolts Subsequently

After 2 washers and 3 nuts have been drawn, the box now contains 4 items, all of which are bolts. We then draw these 4 bolts one after another without replacement.

• Number of items remaining after 2 washers and 3 nuts are drawn: \(9 - 2 - 3 = 4\)
• Number of bolts remaining: 4
• Probability of drawing the first bolt: $$P(\text{1st bolt}) = \frac{4}{4} = 1$$
• Probability of drawing the second bolt: $$P(\text{2nd bolt}) = \frac{3}{3} = 1$$
• Probability of drawing the third bolt: $$P(\text{3rd bolt}) = \frac{2}{2} = 1$$
• Probability of drawing the fourth bolt: $$P(\text{4th bolt}) = \frac{1}{1} = 1$$
• The combined probability of drawing 4 bolts subsequently is: $$P(\text{4 bolts subsequently}) = 1 \times 1 \times 1 \times 1 = 1$$

Calculating the Total Probability of the Sequence

To find the overall probability of this specific sequence (2 washers, then 3 nuts, then 4 bolts), we multiply the probabilities of each stage:

$$P(\text{Sequence}) = P(\text{2 washers first}) \times P(\text{3 nuts subsequently}) \times P(\text{4 bolts subsequently})$$

$$P(\text{Sequence}) = \frac{1}{36} \times \frac{1}{35} \times 1$$

$$P(\text{Sequence}) = \frac{1}{36 \times 35}$$

$$P(\text{Sequence}) = \frac{1}{1260}$$

Therefore, the probability of drawing 2 washers first, followed by 3 nuts, and subsequently the 4 bolts is \(\frac{1}{1260}\).