Modern Math - Probability - Previous Year CAT/MBA Questions

Answer: Option A

Video Explanation

Text Explanation :

Required probability = $\frac{\mathrm{No}. \mathrm{of} \mathrm{ways} \mathrm{of} \mathrm{selecting} 2 \mathrm{smaller} \mathrm{squares} \mathrm{with} \mathrm{having} \mathrm{common} \mathrm{side}}{\mathrm{No}. \mathrm{of} \mathrm{ways} \mathrm{of} \mathrm{selecting} 2 \mathrm{smaller} \mathrm{squares}}$

No. of ways of selecting 2 smaller squares = ⁶⁴C₂ = 32 × 63

Now, to select 2 squares with common side, there are two cases possible.

Case 1: Horizontal pair is selected.
In the first row 2 squares can be selected in 7 ways i.e., (R₁C₁, R₂C₂), (R₁C₂, R₂C₃), ..., (R₁C₇, R₂C₈)
Similarly 2 squares can be selected from each row in 7 ways.
∴ No. of ways of selecting 2 smaller horizontal pair of squares = 7 × 8 = 56.

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Case 2: Vertical pair is selected.
In the first column 2 squares can be selected in 7 ways i.e., (C₁R₁, C₁R₂), (C₁R₂, C₁R₃), ..., (C₁R₇, C₁R₈)
Similarly 2 squares can be selected from each column in 7 ways.
∴ No. of ways of selecting 2 smaller vertical pair of squares = 7 × 8 = 56.

⇒ Total no. of ways of selecting 2 smaller squares with common side = 56 + 56 = 112

∴ Required probability = $\frac{112}{32\times 63}$ = $\frac{112}{2016}$

Hence, option (a).

Answer: Option B

Text Explanation :

The total number of ways of selecting 3 notes from the :

five 10-rupee notes, three 20-rupee notes, and two 50-rupee notes = 10 notes in total.

10C₃​ ​​​​​​= 120

The possibilities for the value of the three notes combined is at least  90 :

Rs 50 + Rs 20 + Rs 20 : The possibilities for this selection is:

2C₁ . 3C₂. Selection of one Rs 50 note from the two and selection of 2 Rs 20 notes from the three.

Rs 50 + Rs 50 + Rs 10:

(2C₂) . (5C₁) : Selection of two Rs 50 notes from the two and selection of 1 Rs 10 notes from the five.

Rs 50 + Rs 50 + Rs 20:

(2C₂) . (3C₁): Selection of two Rs 50 notes from the two and selection of 1 Rs 20 notes from the three.

A total of 6 + 5 + 3 = 14 possibilties

The probability is $\frac{14}{120}$ = 7/60.

Answer: Option C

Text Explanation :

Number of white phones = 2
Number of black phones = 2
Number of red phones = 1
customer 1 will have 3 choices
customer 2 will have 3 choices
customer 3 will have 3 choices
Hence total choices = 3 x 3 x 3 = 27
The cases not possible = BBB, RRR,WWW, RRB,RBR,BRR, RRW,RWR, WRR
Possible cases = 18
Probability = 18/27 = 2/3

Answer: Option B

Text Explanation :

Sum of 3rd row = sum of 2nd column
⇒ 2x + 4y = y + 2z - 1
⇒ 2x + 3y - 2z =  -1               ...(A)
Sum of diagonals are also equal
⇒ 3x + 4y + z - 1 = y + z + 2x + y + z
⇒ x + 2y - z = 1                     ...(B)
Solving A and B we get y = 3
Putting it in A, we get x - z = -5 ...(C)
Sum of 1st row = sum of 2nd column
5x + 5y + z = 3x + 4y + 2z
⇒ 2x + y - z = 0
Since y = 3, 2x - z = -3              ...(D)
Solving C and D we get x = 2 and z = 7
Hence N = 36

Answer: Option E

Text Explanation :

Let the 6 digit number be _ _ _ 42_
It is divisible by 2,3,5,7,11,13

Hence, the number is a multiple of 2 × 3 × 5 × 7 × 11 × 13 = 30030

Now, 30030 × k = _ _ _ 4 2 _

Last digit would definitely be 0.

Second last digits which is 2 should be the last digit of 3 × k. Hence, last digit of k should be 4.
​​​​​​​∴ k = 4 or 14 or 24 and so on.

The only possible value of k from the one's given above satisying the conditions given is 14.

​​​​​​​∴ 30030 × 14 = 420420

​​​​​​​∴ The first digit is 4.

Hence, option (e).

Answer: Option B

Video Explanation

Text Explanation :

The number of ways of selecting one card from Ashok's bag and other from Shilpa’s bag = ⁴⁰C₁ × ⁵C₁ = 200 ways

Favourable cases:

Cards 1 or 2 or 3: If card 1 or 2 or 3 is drawn from Shilpa’s bad remainder in each case will be less than or equal to 2.

∴ Favourable cases = ⁴⁰C₁ × ³C₁ = 120 ways

Card 4: If card 4 is drawn from Shilpa’s bag we need to eliminate those cases when the remainder will be 3.

⇒ We need to eliminate those cards from Ashok’s bag which have number of the form 4x + 3.

∴ Cards of the form 4x + 3 are 3, 7, 11, …., 39 i.e., 10 cards.

∴ Favourable cases = (40 – 10) × 1 = 30 ways

Card 5: If card 5 is drawn from Shilpa’s bag we need to eliminate those cases when the remainder will be 3 or 4.

⇒ We need to eliminate those cards from Ashok’s bag which have number of the form 5x + 3 or 5x + 4.

∴ Cards of the form 5x + 3 are 3, 8, 13, …., 38 i.e., 8 cards.
∴ Cards of the form 5x + 4 are 4, 9, 14, …., 39 i.e., 8 cards.

∴ Favourable cases = (40 - 16) × 1 = 24 ways

∴ Total Favourable cases = 120 + 30 + 24 = 174 ways

⇒ Probability of remainder not greater than 2 = 174/200 = 0.87

Hence, option (b).

Answer: Option E

Text Explanation :

Let’s consider one of the tiles with side 10 cm. For the coin to land completely within the tile, its centre should fall anywhere inside the square of side 4 cm.

​​​​​​​

∴ Favourable area = 4 × 4 = 16

Total area available = 10 × 10 = 100

∴ Required probability = 0.16

Hence, option (e).

Answer: Option C

Text Explanation :

Total combinations when a die is rolled twice = 6 × 6 = 36

Case 1: The first roll = 1
The second roll = (2, 3, 4, 5, 6)
∴ Favourable outcomes = 5

Case 2: The first roll = 2
The second roll = (3, 4, 5, 6)
∴ Favourable outcomes = 4

Case 3: The first roll = 3
The second roll = (4, 5, 6)
∴ Favourable outcomes = 3

Case 4: The first roll = 4
The second roll = (5, 6)
∴ Favourable outcomes = 2

Case 5: The first roll = 5
The second roll = (6)
∴ Favourable outcomes = 1

Hence, the number of favourable outcomes = 5 + 4 + 3 + 2 + 1 = 15

Therefore, the probability = 15/36 

Alternately,
Let us calculate the number of ways a different number comes up both the die = 6 × 5 = 30.

Out of these 30 ways, in half of these first die will have higher number and in other half second die will have higher number.

∴ Probability that second die has higher number = 15/36.

Hence, option (c).

Answer: Option E

Video Explanation

Text Explanation :

Required Probability = 1 – P(receiving no gift)

P(receiving no gift) = 0.4 × 0.2 × 0.1 × 0.5

= 0.0040

1 – P(receiving no gift) = 0.996

Hence, option (e).