CSAT Mixed Practice: Work Rates and Probability

Work rates

Treat one complete job as 1 unit. If A finishes it alone in 12 days, A's rate is 1/12 job per day. If B takes 18 days, B's rate is 1/18 job per day. If both start together on the whole job and maintain these rates without interference, their combined rate is 1/12 + 1/18 = 5/36 job per day. Time = work / rate = 1 ÷ (5/36) = 36/5 = 7.2 days. Completion times are not added because both people work during the same elapsed time.

For a filling pipe, use a positive rate; for a draining pipe, use a negative rate. Divide the amount still to be filled by the net rate only when that rate is positive and constant over the interval. A zero or negative net rate cannot fill an initially empty tank in this model.

Probability from equally likely outcomes

For a finite set of equally likely outcomes, probability = favourable outcomes / total outcomes. For a fair die, the even outcomes are 2, 4 and 6, so P(even) = 3/6 = 1/2. This counting formula cannot be used merely by counting possible outcomes when those outcomes have different probabilities.

A bag contains 3 red and 2 blue balls. If every remaining ball has the same chance of being drawn at each draw, P(red on the first draw) = 3/5. For two draws without replacement, a first red leaves 2 red balls among 4 balls. Thus P(both red) = (3/5)×(2/4) = 3/10. With replacement and a fresh independent uniform draw after mixing, the second red probability is again 3/5, giving (3/5)×(3/5) = 9/25. Replacement restores the counts; the stated independent draw makes the multiplication valid in this form.

At least one

For two independent fair coin tosses, the equally likely ordered outcomes are HH, HT, TH and TT, where H means heads and T means tails. Only TT has no heads. Therefore P(at least one head) = 1 − P(no heads) = 1 − 1/4 = 3/4. The complement includes exactly the outcomes excluded by the requested event.

Mixed worked drill

1. A alone takes 10 days and B alone takes 15 days. Both start the whole job together and work at constant additive rates. Their rate is 1/10 + 1/15 = 1/6 job per day, so the time is 1 ÷ (1/6) = 6 days.
2. A pipe fills an empty tank in 8 hours; another empties a full tank in 12 hours. Starting empty, both are opened together. Assume the stated rates remain constant whenever water is available to drain. Net rate = 1/8 − 1/12 = 1/24 tank per hour. Filling time = 1 ÷ (1/24) = 24 hours. This calculation models fixed rates, not a flow rate that changes with water level.
3. One card is selected uniformly from cards numbered 1–10. The multiples of 3 are 3, 6 and 9. There are 3 favourable cards among 10 cards, so the probability is 3/10.
4. Two independent fair dice are rolled. The ordered outcomes with sum 7 are (1,6), (2,5), (3,4), (4,3), (5,2) and (6,1). There are 6 such outcomes among 36 equally likely ordered outcomes, so the probability is 6/36 = 1/6.
5. A bag has 3 red and 2 blue balls. Two balls are drawn without replacement, each draw uniform among the remaining balls. One of each colour can occur as red then blue or blue then red. These orders cannot both occur in the same two-draw result, so add their probabilities: (3/5×2/4)+(2/5×3/4) = 3/10 + 3/10 = 3/5.

When one worker leaves

A completes a job alone in 12 days and B in 18 days. Both work for 3 days, then A leaves. Assume their rates stay constant and add while both work. Work completed = 3×(1/12 + 1/18) = 5/12. Work left = 1 − 5/12 = 7/12. B needs (7/12) ÷ (1/18) = 10.5 more days. Total elapsed time = 3 + 10.5 = 13.5 days. Using the original combined rate for the remaining work would be wrong because only B is now working.

Overlapping events

On one fair die, find the probability of an even number or a multiple of 3. The even set is {2,4,6}; the multiple-of-3 set is {3,6}. Adding 3/6 + 2/6 counts 6 twice. Subtract that overlap once: P(even or multiple of 3) = 3/6 + 2/6 − 1/6 = 4/6 = 2/3. The union is {2,3,4,6}, which confirms the answer directly. In general, P(A or B) = P(A) + P(B) − P(A and B).