Probability of Other Events

Question 1. A carton consists of 100 shirts of which 88 are good, 8 have minor defects and 4 have major defects. Jimmy, a trader, will only accept the shirts which are good, but Sujatha, another trader, will only reject the shirts which have major defects. One shirt is drawn at random from the carton. What is the probability that
(i) it is acceptable to Jimmy?
(ii) it is acceptable to John?
solution:
One shirt is drawn at random from the carton of 100 shirts. Therefore, there are 100 equally likely outcomes.
(i) The number of outcomes favorable (i.e., acceptable) to Jimmy =88 (Why?)
Therefore, P(shirt is acceptable to Jimmy) =88/100=0.88.
(ii) The number of outcomes favorable to John =88+8=96 (Why?)
So, P(shirt is acceptable to John) =96/100=0.96

Q2. A piggy bank contains hundred 50 p coins, fifty Rs 1 coins, twenty Rs 2 coins and ten Rs 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin
(i) Will be a 50 p coin?
(ii) Will not be a Rs.5 coin?
answer:
Total number of coins in a piggy bank=100+50+20+10=180
(i) Number of 50 p coins=100

Probability of getting a 50 p coin= 100/180=5/9
(ii) Number of Rs 5 coins =10
Probability of getting a Rs 5 coin= 10/180=1/18
Probability of not getting a Rs 5 coin=1-1/18=17/18

Q3. Two customers Sam and Emma are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the
probability that both will visit the shop on
(i) the same day?
(ii) consecutive days?
(iii) different days?
answer:
There are a total of 5 days. Sam can go to the shop in 5 ways and Emma can go to the shop in 5 ways.
Therefore, total number of outcomes =5×5=25 (i) They can reach on the same day in 5 ways.
i.e., (t, t), (w, w), (th, th), (f, f), (s, s)
P(both will reach on same day)=5/25=⅕
(ii) They can reach on consecutive days in these 8 ways – (t, w), (w, th), (th, f), (f, s), (w, t), (th, w), (f, th), (s, f).
Therefore, P(both will reach on consecutive days)=8/25
(iii) P(both will reach on same day)=⅕ [(From (i)]

P(both will reach on different days)=1-⅕=⅘

Q4. On her vacations Veena visits four cities (A, B, C and D) in a random order. What is the probability that she visits
(i). A before B? (ii). A before B and B before C?
solution:
The number of arrangements (orders) in which Veena can visit four cities A, B, C, or D is 4! i.e., 24.Therefore, n(S)=24.
Since the number of elements in the sample space of the experiment is 24 all of these outcomes are considered to be equally likely. A sample space for the experiment is
S={ABCD, ABDC, ACBD, ACDB, ADBC, ADCB, BACD, BADC, BDAC, BDCA, BCAD, BCDA, CABD, CADB, CBDA, CBAD, CDAB, CDBA. DABC, DACB, DBCA, DBAC, DCAB, DCBA}

(i). Let the event ‘she visits A before B’ be denoted by E. Therefore,
E={ABCD, CAM, DABC, ABDC, CADB, DACB, ACBD, ACDB, ADBC, CDAB, DCAB, ADCB}. Thus

(ii). Let the event ‘Veena visits A before B and B before C’ he denoted by F. Here
F={ABCD, DABC, ABDC, ADBC}
Therefore,

Let’s read about factorial.

Q5. Five letters were to be put in five corresponding addressed envelopes. If each letter was put in one of the envelopes then what is the probability that each letter will be put in correct envelope?
a. 1/2⁵ b. 1/5!
c. 1/⁵C₂ d. None of the above
correct: b, solution:
First letter can be put in 5 ways. Second letter can be put in 4 ways. And so on last letter can be put in only 1 way.
So number of ways all letters can be put =5!. Probability that all the Ietters are put on correct envelope =1/5!.

Q6. Three letters are dictated to three persons and an envelope is addressed to each of them, the letters are inserted into the envelopes at random so that each envelope contains exactly one letter. Find the probability that at least one letter is in its proper envelope.
solution:
Three events that one letter isin its proper envelope. No event that two letters are in their respective proper envelopes. One event that three letters are in their respective proper envelopes. The sample space contains ³P₃=3!. Using permutation because of regarding of sequence.
Let L₁, L₂, L₃ be three letters and E₁, E₂, E₃ be their corresponding envelopes respectively. A pair is prover, if there is a L_x E_x {x=1,2,3}.