JEE Advanced (Single Correct MCQs): Probability - EmSAT Achieve MCQ

If two dice are tossed, the total number of sample space = 36

P(A)= Probability that the first die shows an even number = 18/36 = 1/2

P(B) = Probability that the second die shows an odd number = 18/36 = ½

P(A∩B) The probability of even on first die and odd on second die = 9/36 = ¼

P(A∩B) = ¼, and hence they are not mutually exclusive.

Also, P(A). P(B) = (½).(½) = ¼

Thus, P(A∩B) = P(A).P(B)

Hence, they are independent.

Hence, Option (4) is the correct answer.

P (A ∪ B) = P (A) + P (B) – P (A ∩ B) = 0.25 + 0.50 – 0.14 = 0.61
∴ P (A' ∩ B') = P ((A ∪ B)') = 1 – P (A ∪ B)
= 1– 0.61 = 0.39

p = 0.4, n = 3, P (X ≥ 1) = ? ⇒ q = 0.6
∴ P (X ≥ 1) = 1– P (X = 0)
                  = 1– 3C₀(0.4)⁰ (0.6)³ = 1 – 0.216 = 0.784

n = 7

Prob. of getting any no. out 1, 2, 3, … 9 is p = 9/15

∴ q = 6/5

         [Binomial distribution]

Favourable cases = 6; {(1, 1, 1), (2, 2, 2),…(6, 6, 6)}

Total cases  = 6 × 6 × 6 = 216
∴Req. prob. = 

Prob. of a getting a white ball in a single draw

Prob. of getting a white ball 4th time in the 7th draw = P (getting 3 W in 6 draws) × P (getting W ball at 7th draw)

Prob. of one coin showing head = p
∴ Prob of one coin showing tail = 1– p ATQ coin is tossed 100 times and prob. of 50 coins showing head = prob of 51 coins showing head.
Using binomial prob. distribution

P (X = r)= ⁿC_r p^rq^n-r ,

we get,    

⇒ 51 – 51 p = 50 p

⇒ 101 p = 51 ⇒ 

P (at least 7 pts) = P (7pts) + P (8 pts) [∵ At most 8 pts can be scored.]
Now 7 pts can be scored by scoring 2 pts in 3 matches and 1 pt. in one match, similarly 8 pts can be scored by scoring 2 pts in each of the 4 matches.
∴ Req. prob. = ⁴C₁ × [P (2 pts)]³ P (1pt) + [P (2 pts)]⁴ = 4 (0.5)³ × 0.05 + (0.50)⁴  = 0.0250 + 0.0625 = 0.0875

The min. face value is not less than 2 and max. face value is not greater than 5 if we get any of the numbers 2, 3, 4, 5, while total possible out comes are 1, 2 , 3, 4, 5 and 6.

∴ In one thrown of die, prob. of getting any no.
Out of 2, 3, 4 and 5 is 

If the die is rolled four times, then all these events being independent, the required prob.  

P [A ∩ (B ∪ C)] = P [(A ∩ B) ∪ (A ∩ C)] = P (A ∩ B) + P (A ∩ C) – P (A ∩ B ∩ C)
= P (A)  P (B) + P (A)  P (C) – P (A)  P (B) P(C)
= P (A) [P (B) + P (C) – P (B ∩ C)] = P (A) P (B ∪ C)

∴ S₁ is true.
P (A ∩ (B ∩ C)) = P (A) P(B)  P (C) = P (A) P (B ∩ C)
∴ S₂ is also true.

Given that P (India wins) = p = 1/2
∴ P (India loses) = p' = 1/2 Out of 5 matches india’s second win occurs at third test
⇒ India wins third test and simultaneously it has won one match from first two and lost the other.
∴ Required prob. = P (LWW) + P (WLW)

Out of 6 vertices 3 can be chosen in ⁶C₃ ways.
Δ will be equilateral if it is Δ ACE or ΔBDF (2 ways)

∴ Required prob. = 

We know that P (exactly one of A or B occurs)  = P (A) + P (B) – 2P (A ∩ B).
Therefore, P (A) + P (B) – 2P (A ∩ B) = p … (1)
Similarly,  P (B) + P (C) – 2P (B ∩ C) = p … (2)  
and P (C) + P (A) – 2P (C ∩ A) = p …(3)
Adding (1), (2) and (3) we get 2 [P (A) + P (B) + P (C) – P (A ∩ B)
– P (B ∩ C) – P (C ∩ A)] = 3p ⇒ P (A) + P (B) + P (C) – P (A ∩ B)
        – P (B ∩ C) – P (C ∩ A) = 3p/2 … (4)
We are also given that, P (A ∩ B ∩ C) = p² … (5)
Now, P (at least one of A, B and C) = P (A) + P (B) + P (C) – P (A ∩ B) – P (B ∩ C)
– P (C ∩ A) + P (A ∩ B ∩ C)

[using (4) and (5)] 

We know that, 71= 7, 72 = 49, 73 = 343, 74 = 2401, 75 = 16807
∴ 7^k (where k ∈ Z), results in a number whose unit’ss digit is 7 or 9 or 3 or 1.
Now, 7^m + 7ⁿ will be divisible by 5 if unit’s place digit of resulting number is 5 or 0 clearly it can never be 5.
But it can be 0 if we consider values of m and n such that the sum of unit’s place digits become 0. And this can be done by choosing

(25 options each) [7 + 3 = 10]

(25 options each) [9 + 1= 10]

Case I : Thus m can be chosen in 25 ways and n can be chosen in 25 ways

Case II : m can be chosen in 25 ways and n can be chosen in 25 ways

∴ Total no. of selections of m, n so that 7^m + 7ⁿ is divisible by 5 = (25 × 25 + 25 × 25 ) × 2

Note we can interchange values of m and n.
Also no. of total possible selections of m and n out of 100 = 100 × 100

The minimum of two numbers will be less than 4 if at least one of the numbers is less than 4.

∴ P (at least one no. is < 4), = 1– P (both the no’s are ≥ 4)

Given that  

From venn diagram, we see

If a no. is to be divisible by both 2 and 3. It should be divisible by their L.C.M.
∴ L.C.M. of (2 and 3) = 6
∴ Numbers are  =  6, 12, 18 …  96.
Total numbers are = 16

∴ Probability =

In single throw of a dice, probability of getting 1 is =  and prob. of not getting 1 is 

Then getting 1 in even no. of chances = getting 1 in 2nd chance or in 4th chance or in 6th chance and so on

∴ Req. Prob 

Let E₁ ≡ The Indian man is seated adjacent to his wife.
E₂ ≡ Each American man i s seated adja cen t to his wife.

Then P(E₁ ∩ E₂) = 

Now E₁ ∩E₂≡ All men are seated adjacent to their wives.
∴ We can consider the 5 couples as single-single objects which can be arranged in a circle in 4! ways.
But for each couple, husband and wife can interchange their places in 2! ways.
∴ Number of ways when all men are seated adjacent to their wives = 4! × (2!)⁵
Also in all 10 persons can be seated in a circle in 9! ways.

∴

Similarly if each American man is seated adjacent to his wife, considering each American couple as single object and Indian woman and man as seperate objects there are 6 different objects which can be arranged in a circle in 5! ways. Also for each American couple, husband and wife can interchange their places in 2 ! ways.
So the number of ways in which each American man is seated adjacent to his wife.

P( E^c ∩ F^c /G) = P( E ∪ F )^c /G) 1 - P(E∪ F /G)
= 1 - P(E / G) - P(F/ G) + P(E∩ F / G)
= 1 - P(E ) -P(F)+O
(∵ E, F, G are pairwise independent and
P(E ∩F ∩G)=0
⇒ P(E ).P(F )=0 as P(G) >0 ) = P(E^c )- P(F)

We have n (S) = 10, n(A) = 4
Let n(B) = x and  n(A ∩ B)  = y
Then for A and B to be independent events P(A ∩ B) = P(A) P(B)

⇒ y can be 2 or 4 so that x = 5 or 10
∴ n (B) = 5 or 10

If ω is a complex cube root of unity then, we know that ω^3m +ω³ⁿ⁺¹ +ω^{3p +2} = 0 where m, n, p are integers.
∴ r₁, r₂, r₃ should be of the form 3m, 3n + 1 and 3p + 2 taken in any or d er. As r₁ ,r₂ ,r₃ are the numbers obtained on die, these can take any value from 1 to 6.
∴ m can take values 1 or 2, n can take values 0 or 1p can take values 0 or 1
∴ Number of ways of selecting r₁, r₂, r₃ 
= ²C₁ x ²C₁ x ²C₁ x 3!.
Also the total number of ways of getting r₁, r₂, r₃ on die = 6 × 6 × 6

∴ Required probability=

  Let G ≡ oreginal signal is green ⇒ P (G )= 45
E₁ ≡ A receives the signal correctly P(E₁) = 34
E₂ ≡ B receives the signal correctly P(E₂) = 34
E ≡ Signal received by B is green.
Then E can happen in the following ways

D₄ can show an umber appearing on one of D₁, D₂ and D₃ in the following cases.
Case I : D₄ shows a number which is shown by only one of D₁, D₂ and D₃.
D₄ shows a number in ⁶C₁ ways.
One out of D₁, D2 and D3 can be selected in ³C₁ ways.
The selected die shows the same number as on D4 in one way and rest two dice show the different number in 5 ways each.
∴ Number of ways to happen case I = ⁶C₁ × ³C₁ × 1 × 5 × 5 = 450
Case II : D₄ shows a number which is shown by only two of D₁, D₂ and D₃.
As discussed in case I, it can happen in the following number of ways = ⁶C₁ × ³C₂ × 1 × 1 × 5 = 90
Case III : D₄ shows a number which is shown by all three dice D₁, D₂ and D₃.
Number of ways it can be done = ⁶C₁ × ³C₃ × 1 × 1 × 1 = 6
∴ Total number of favourable ways = 450 + 90 + 6 = 546

Also total ways = 6 × 6 × 6 × 6

∴ Required Probability =  

According to given condition , we can h ave the following cases
(I) G G B B B (II) B G G B B
(III) G B G B B (IV) B G B G B
(V) G B B G B
i.e., the two girls can occupy two of the first three places (case I, II, III) or second and fourth (case IV) or first and fourth (case V) places.
Thus favourable cases are = 3 × 2! × 3! + 2 × 2! × 3! = 60 Total ways in which 5 persons can be seated = 5! = 120
∴ Required probability 

Also P(D) =