**Solved Examples on Probability**

**Example 1: **A pair of dice is thrown. Find the probability of obtaining a sum of 8 or getting an even number on both the dice.

**Solution:** Let the events be defined as:

A : obtaining a sum of 8

B : getting an even number on both dice

We are required to find out the total Probability A and B, i.e.

P(AUB) = P(A) + P(B) - P(A∩B)

Now cases favourable to A are (3, 5) (5, 3) (2, 6) (6, 2) (4, 4)

So, P(A) = 5/36

Cases favourable to B : (2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6).

P(B) = 9/36

Now, (2, 6) (6, 2) and (4, 4) are common to both events A and B

So P(A∩B) = 3/36

⇒ P(AUB) = 5/36 + 9/36 - 3/36 = 11/36.

**Example 2:** A number x is selected from first 100 natural numbers. Find the probability that x satisfies the condition x + 100/x > 50.

**Solution:** Total number of ways of selecting x is 100.

Now the given condition is x + 100/x > 50, on analyzing this equation, carefully, we see that this equation is satisfied for all the numbers x such that x > 48 and also for x = 1 and 2

So, favourable number cases is 55

Therefore, probability = 55/100 = 11/20

**Example 3:** 'A' has three share in a lottery in which there are 3 prizes and 6 blanks' 'B' has one share in a lottery in which there is 1 prize and 2 blanks. Show that A's chance of winning a prize of B's chance of winning a prizes in ratio 16:7.

**Solution:** We can solve this question by two methods:

**Method 1:**

'A' may win one, two or all the three prizes, the total probability is = (^{3}C_{1}^{6}C_{2})/(^{9}C_{3}) + (^{3}C_{2}^{6}C_{1})/(9C_{3}) + (^{3}C_{3}^{6}C_{0})/(^{9}C_{3}) = (45 + 18 + 1)/84 = 64/84 = 16/21

P(B) = (^{1}C_{1})/(^{3}C_{1}) = 1/3

(P(A))/(P(B)) = (16×3)/21 = 16/7

⇒ P(A) : P(B) = 16 : 7 (Proved)

**Method 2:**

A can win one, two or all the prizes in the following manner

1 prize, 2 blanks 2 prizes, 1 blank 3 prizes

I II III

All the three cases where 'A' can win are independent and mutually exclusive. Hence probability that A can win prize is given as

P(A) = P(I) + P(II) + P(III)

Let's calculate P(I): It is something like this:

A bag contains 3 white balls (prizes) and 6 black balls (blanks) then what is the probability that out of 3 random draws, there is 1 white ball (prize) and 2 black balls (blanks).

Which one can calculate very easily.

Favourable ways for this event =^{3}C_{1 }× ^{6}C_{2}

Total number of ways =^{ 9}C_{3} P(I) = (^{3}C_{1} ×^{ 6}C_{2})/(^{9}C_{3}),

Similarly we can calculate P(II) and P(III)

Hence, P(A) = (^{3}C_{1} × ^{6}C_{2})/(^{9}C_{3}) + (^{3}C_{2} × ^{6}C_{1})/(^{9}C_{3}) + (^{3}C_{3} × ^{6}C_{0})/(^{9}C_{3}) = 16/21

B has only one share in a lottery, which consists of 1 prize and 2 blanks' therefore B can at most win only one prize. Hence

P(B) = ^{1}C_{1}/^{3}C_{1} = 1/3

(P(A))/(P(B)) = (16 × 3)/21 = 16/7

P(A) : P(B) = 16:7 (Ans.)

**Example 4:** There are three events A, B and C one of which must , and only one can happen, the odd are 8 to 3 against and 2 to 5 for B. Find the odds, against C.

**Solution:** P(A) = 3/11, P(B) = 2/7, P(C) = x (say)

Since at most one and only one can happen therefore A, B, C are mutually exclusive and exhaustive events.

So, P(A) + P(B) + P(C) = 1

⇒ 3/11 + 2/7 + x = 1

x = (77 – 21 – 22)/77 = 34/77

Odds against C = (77 - 34) : 34 or 43 : 34 (Ans.)

**Example 5:** An unbiased coin is tossed. If the result is head, a pair of unbiased dice is rolled and the number obtained by adding the number on the two faces are noted. If the result is a tail, a card from a well-shuffled pack of 11 cards numbered 2, 3, 4.... .., 12 is picked & the number on the card is noted. What is the probability that the number noted is 7 or 8?

**Solution:** Let us define the events

A : head appears.

B : Tail appears

C : 7 or 8 is noted.

We have to find the probability of C i.e. P (C)

P(C) = P(A) P (C/A) + P(B) P(C/B)

Now we calculate each of the constituents one by one

P(A) = probability of appearing head = 1/2

P(C/A) = Probability that event C takes place i.e. 7 or 8 being noted when head has already appeared. (If something has already happened then it becomes certain, i.e. now it is certain that head has appeared we have to certainly roll a pair of unbiased dice).

= 11/36 (since (6, 1) (1, 6) (5, 2) (2, 5) (3, 4) (4, 3) (6, 2) (2, 6) (3, 5) (5, 3) (4, 4) i.e. 11 favourable cases and of course 6 × 6 = 36 total number of cases)

Similarly, P(B) = 1/2

P(B/C) = 2/11 (Two favourable cases (7 and 8) and 11 total number of cases).

Hence, P(C) = ½ × 11/36 + ½ × 2/11 = 193/792 (Ans.)

**Example 6: **If the probability for A to win a game against B is 0.4. If A has an option of playing either a "best of 3 games" or a "best of 5 games" match against B, which option should A choose so that the probability of his winning the match is higher?

**Solution:** Let’s define the events

A : A wins a game against B ⇒ P(A) = 0.4

A^{C}: A losses a game against B ⇒ P(A^{C}) = 0.6

B : A wins match against B

**Case I:**

When A plays a "Best of three games" match.

In this case A will have to win at least two of three games if he has to win the match.

This he can do by winning either 2 games or all of the 3 games in the manner.

A**A ^{C}**A +

**A**AA + AA

^{C}**A**+ AAA

^{C}I II III IV

P(B) =^{ 3}C_{2}A^{2}(A^{C}) +^{3}C_{3}(A)^{3} = 3 × (0.4)^{2} × (0.6) + (0.4)^{3} = 0.352

**Case II:**

When A play a "Best of 5 games against" B. Then

P(B) = ^{5}C_{3}A^{3}(A^{C})^{2} + ^{5}C_{4}(A)^{4} A^{C} +^{ 5}C_{5}(A)^{5}

= 10 × (0.4)^{3} (0.6)^{2 }+ 5 × (0.4)^{4 }(0.6) + (0.4)^{5} = 0.23424

Therefore, A should choose Best of three games (Ans.)

**Example 7: **Sixteen players P_{1}, P_{2}, ….. P_{16} play in a tournament. They are divided into eight pairs at random. From each pair, a winner is decided on the basis of a game played between the two players of the pair. Assuming that all the players are of equal strength, the probability that exactly one of the players P_{1} and P_{2} is among the eight winners is

(a) 4/15

(a) 4/15

(a) 4/15

(a) 4/15

**Solution:** Let E_{1} and E_{2} denote the event that P_{1} and P_{2} are paired or not paired together. Let A denote the event that one of the two players P_{1} and P_{2} is amongst the winners.

Since, P_{1} can be paired with any of the remaining 15 players, so P(E_{1}) = 1/15

and P(E_{2}) = 1 – P(E_{1}) = 1 – 1/15 = 14/15

In case E_{1} occurs, it is certain that one of P_{1} and P_{2} will be among the winners. In case E_{2} occurs, the probability that exactly one of P_{1} and P_{2} is among the winners is

P[(P_{1} ∩ P_{2}^{C}) ∪ (P_{1}^{C} ∩ P_{2})] = P(P_{1} ∩ P_{2}^{C}) + P(P_{1}^{C} ∩ P_{2})

= P(P_{1}) P(P_{2}^{C}) + P(P_{1}^{C}) P(P_{2})

= ½ (1 - 1/2) + (1 - 1/2)1/2

= ¼ + ¼

= ½

i.e. P(A/E_{1}) = 1 and P(A/E_{2}) = ½

By the total probability rule,

P(A) = P(E_{1}) . P(A/E_{1}) + P(E_{2})P(A/E_{2})

= 1/15 (1) + 14/15(1/2)

= 8/15

**Example 8:** A is a set containing n elements. A subset P of A is chosen at random. The set A is reconstructed by replacing the elements of the subset of P. A sub set Q of A is again chosen at random. Find the probability that P and Q have no common elements.

**Solution:** In set P we can have no element i.e.Φ, 1 element, 2 elements, ...... upto n elements. If we have no element in P, we will leave by all the elements and number of set Q formed by those elements will have no common element in common with P. Similarly, it there are r elements in P we are left with rest of (n - r) element to form Q, satisfying the condition that P and Q should be disjoint.

Hence, the total number of ways in which P and Q are disjoint can be given by = ∑^{n} _{r = 0} ^{n}C_{r}(2)^{(n-r)} = (3)^{n}

[Suppose you have a 5 element set {1, 2, 3, 4, 5} and you formed a subset of 2 elements, which you can do in ^{5}C_{2} ways. Let's select 1 and 2, then we are left with 3, 4 and 5.

Now we can form (2)^{3} = 8 i.e. {Φ}, {3}, {4}, {5}, {3, 4}, {3, 5}, {4, 5}, {3, 4, 5} subsets which will have no common element with the previously chosen subset.

So the total number of ways in which you can form two disjoint set is ^{5}C_{0}(2)^{5} + ^{5}C_{1}(2)^{4 }+^{ 5}C_{2}(2)^{3} +^{ 5}C_{3}(2)^{2 }+^{ 5}C_{4}(2)^{1} +^{ 5 }C_{5}(2)^{0} = (3)^{5}]

Total number of ways in which we can form P and Q

= ^{n}C_{0}(2)^{n} + ^{n}C_{1}(2)^{n} +^{ n}C_{2}(2)^{n} + … + ^{n}C_{n}(2)^{n} = (4)^{n}

or simply total number of ways = Total number of ways of forming P × total number of ways of forming Q.= 2^{n}× 2^{n }= (4)^{n}.

So, required probability = (3/4)^{n}.(Ans.)

**Example 9:** In a test an examinee either guesses or copies or knows that answer to a multiple choice question which has 4 choices. The probability that he makes a guess is 1/3 and the probability that he copies is 1/6. The probability that his answer is correct, given the copied it, is 1/8. Find the probability that he knew the answer to the question, given that he answered it correctly.

**Solution:** P(g) = probability of guessing = 1/3

P(c) = probability of copying = 1/6

P(k) = probability of knowing = 1 - 1/3 - 1/6 = 1/2

(Since the three-event g, c and k are mutually exclusive and exhaustive)

P(w) = probability that answer is correct

P(k/w) = (P(w/k).P(k))/(P(w/c)P(c) + P(w/k)P(k) + P(w/g)P (g)) (using Baye's theorem)

= (1×1/2)/((1/8,1/6) + (1×1/2) + (1/4×1/3) )

= 24/29(Ans.)

**Example 10:** A speaks truth 3 out of 4 times. He reported that Mohan Bagan has won the match. Find the probability that his report was correct.

**Solution:**

**Method 1:**

Let T : A speaks the truth

B : Mohan Bagan won the match

Given, P(T) = 3/4

.· . P(T^{C}) = 1 - 1/3 = 1/4

A match can be won, drawn or loosen

.·. P(B/T) = 1/3 P(B/T^{C}) = 2/3.

Using Baye's theorem we get

P(T/B) = (P(T).P(B/T))/(P(T).P(B/T) + P(T^{C})P(B/T^{C}))

= 3/4×1/3)/(3/4×1/3 + 1/4×2/3) = (1/4)/(5/12)

=3/5

**Method 2:**

Let, T : The man speaks truth

A : Mohan Bagan won the match

B : He reported that Mohan Bagan has won.

P(A) = 1/3(the match may also end in a draw)

P(T) = 3/4

P(B) = P(A) P(T) + P(A^{C}) P(T^{C})

= 1/3×3/4 + 2/3×1/4

= ¼ + 1/6

= (3+2)/12 = 5/12

P(T/B) = (P(B/T).P(T))/(P(B)) = (1/3×3/4)/(5/12)

= 3/5 (Ans.)

**Example 11:** A lot contains 50 defective and 50 non-defective bulbs. Two bulbs are drawn at random one at a time with replacement. The event A, B, C are defined as:

A : the first bulb is defective

B : the second bulb is non-defective

C : both the bulbs are defective or both are non-defective,

Determine whether

(i) A, B, C are pair wise independent.

(ii) A, B, c are mutually independent.

**Solution:** Let probability of defective bulb is denoted by P_{d} and non-defective bulb is denoted by P_{n}.

So, P_{d} = P_{n} = ½

Let, D_{1}: First bulb is defective

D_{2}: Second bulb is defective

N_{1}: First bulb is non-defective

N_{2}: Second bulb is non-defective

Now event A is First bulb is defective and second bulb is defective or First bulb is defective and second bulb is non-defective i.e.

A is D_{1}D_{2}, D_{1}N_{2}

B is D_{1}N_{2}, N_{1}N_{2} and

C is D_{1}D_{2}, N_{1}N_{2}

⇒ A∩B = D_{1}N_{2}

B∩C = N_{1}N_{2}

C∩A = D_{1}D_{2}

⇒ P(A∩B) = 1/4 P(B∩C) = P(C∩A)

also P(A) = 1/2*1/2 + 1/2*1/2 = 1/2 = P(B) = P(C)

⇒ P(A ∩ B) = P(A) P(B)

P(B ∩ C) = P(B) P(C)

P(C ∩ A) = P(C) P(A)

This condition is insufficient for A, B, C to be pair wise independent

Now, A ∩ B ∩ C = Φ ⇒ P(A ∩ B ∩ C) ≠ P(A) P(B) P(C)

Since, there is no element, which is common to all A, B, C.

So, A, B, C are not mutually independent events.(Ans.)

**Example 12:** A sample consists of integer 1, 2......... 2n. The probability of choosing the integer k is proportional to log k. Find the conditional probability of choosing the integer 2 given that an even integer is chosen.

**Solution:** P(integer k is chosen)k = C log k (·.· it is proportional to log k), where C is some constant.

Now let even A : 2 is chosen

B : An even integer is chosen, we have to find conditional probability of A given event B has happened i.e. P(A/B)

P(A/B) = (P(A∩B))/(P(B)) = (P(A).P(B/A))/(ΣP(A)P(B/A)) = (C.log2)/(C log2+C log4+C log6+.........+ C log 2n)

= (log2)/log(2^{n}.(|n)) = (log2)/(n log2 + log(|n)) (Ans)

**Example 13:** Two players A and B toss a coin alternatively, with A beginning the game. The players who first throw a head is deemed to be the winter. B's coin is fair and A's is biased and has a probability p showing a head. Find the value of p so that the gam e is equi-probable to both the players.

**Solution:** Player 'A' wins if he gets head in the first trial or in third [If B does not get head in his first trial] and so on.

P(A) = p + (1 - p) × 1/2 × p + (1 - p) × 1/2 × (1 - p) × 1/2 × p +...∞

This is an infinite G.P. of first term p and common ratio (1-p)/2 = p/(1 – ((1 - p)/2)) = (2p)/(p + 1)

According to given condition

P(A) = P(B)

⇒ 2p/(1+p) = 1 – 2p/(p+1)

⇒ (4p)/(1+p) = 1

⇒ p = 1/3 (Ans.)

**Example 14:** In a multiple choice question, there are four alternative answers, of which one or more are correct. A candidate will get marks in the question only when if he ticks all the correct answers. The candidate decides to tick answers at random. If he is allowed up to three chances to answer the question, find the probability that he will get marks in the question.

**Solution:** Let the multiple choice question has four alternatives in the form

(a) _____________ (b) _____________

(c) _____________ (d) _____________

The possible answers are a, b, c, d, ab, ac, ad, bc, bd, cd, abc, abd, acd, bcd, abcd

Hence the total number of possible answers of such questions = 15.

If a student is given 3 chances to make guess, he will definitely choose 3 different answers.

Hence number of favourable way = 3.

Therefore, Probability = 3/15 = 1/5 (Ans.)

**Alternative Method:**

If you are bit practiced you can nearly visualize that there are 4 elements (options) to form number of subsets (possible answers), which we can do in 2^{4}- 1 = 15 ways. (1 is subtracted, as null set in this case means you are not choosing any answer at all).

Total number of favourable way = 3

Hence probability = 3/15 = 1/5(Ans.)

**Example 15:** Four cards are drawn at random from a well-shuffled pack of 52 cards. Find the probability that there is exactly one pair.

**Solution:** Let H, C, D, S denotes heart, club, diamond and spade respectively. We have to draw 4 cards at random, so that it consists of exactly one pair. A pair means two cards of same denomination. i.e. (5H, 5C) or (6C, 6 D), so any one of such (i.e. having exactly one pair) out of four-selected card should look like. 2H, 2S, 3C, 5D

Now we will find in how many ways, we can do this.

We have a pack of card like this.

1 H, 1 C, 1 D, 1 S ........ (1)

2 H, 2 C, 2 D, 2 S ...... (2)

3 H, 3 C, 3 D, 3 S ...... (3)

….….….….….….….

J H, J C, J D, J S ...... (11)

Q H, Q C, Q D, Q S ...... (12)

K H, K C, K D, K S ...... (13)

Firstly, in how many ways can we select a pair? There are 13 rows in above depiction of a pack of card. Each row consists of 4 cards of same denomination. If we select any of the row, which we can do in ^{13}C_{1} ways (say we have selected 3 row) and in that selected row if we select any of the 2 cards (say 3 H and 3 C) out of 4 which we can do in ^{4}C_{2}ways, we will get total number of selecting a pair.

i.e. ^{13}C_{1} × ^{4}C_{2}

Having selected pair we have to select two cards of two different denominations.

Now we are left with 50 cards but out of those 50 cards 2 cards (in our case 3 D and 3 S) are of no use as they will destroy our pair (they will make either triplet or quartet). Hence we are left with only 48 cards.

Since we have to select 2 more cards and if one says it can be done in ^{48}C_{2} ways. It will be wrong. In our case, as those ^{48}C_{2} combinations will have many pairs included in itself, and we need only one pair. So what we need is to select two cards carefully. We will do it one by one.

We will select one card, which we can do in ^{48}C_{1} ways (say we have selected 5 S).

Number of ways in which we have selected 3 cards

= ^{13}C_{1} ×^{ 4}C_{2} × ^{48}C_{1}.

Now we are left with 47 cards but out of this 47 cards there are 3 cards (5 H, 5 C, 5 D) are of no use as selection of any one of them will form a pair. Therefore, we are left with only 44 cards.

We can select any of one card without violating any of our conditions in ^{44}C_{1} ways.

Hence favourable numbers of ways = ^{13}C_{1} × ^{4}C_{2} × ^{48}C_{1} × ^{44}C_{1}

Total numbers of ways = ^{52}C_{4}

Probability = (^{13}C_{1} × ^{4}C_{2} × ^{48}C_{1} ×^{ 44}C_{1}) / (^{52}C_{4}) (Ans.)

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