# PE 1 - Probability | Modern Math - Probability

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**PE 1 - Probability | Modern Math - Probability**

If 3 consecutive letters are selected at random from the English alphabet, then the probability that one of the letters is a vowel is:

- (a)
7/12

- (b)
5/12

- (c)
5/8

- (d)
None of these

Answer: Option D

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**Explanation** :

**A** B C D

**F G H**

__E__**J K L M N**

__I__**P Q R S T**

__O__**V W X Y Z**

__U__There are a total of (26 – 2 =) 24 sets of 3 consecutive letters in the alphabet series.

Out these 3 consecutive letters which don’t contain vowels are:

B C D

F G H

J K L

K L M

L M N

P Q R

Q R S

R S T

V W X

W X Y

X Y Z

i.e., total 11 sets.

⇒ Favourable sets = 24 – 11 = 13

∴ Required probability = 13/24.

Hence, option (d).

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**PE 1 - Probability | Modern Math - Probability**

Box A contains 4 red and 6 green balls. Box B contains 7 red and 3 green balls. A ball is drawn from box A and without seeing its colour, it is put into box B. If a ball is now drawn from box B, then the probability that it is red is:

- (a)
6/11

- (b)
43/55

- (c)
37/55

- (d)
49/5

Answer: Option C

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**Explanation** :

**Case 1**: A red ball is transferred from A → B.

P(red ball A → B) = 4/10 = 2/5

Now, box B has 8 red and 3 green balls.

P(red ball is picked from B given that red A → B) = 8/11

∴ P(red ball is picked from B) = 2/5 × 8/11 = 16/55

**Case 2**: A green ball is transferred from A → B.

P(green ball A → B) = 6/10 = 3/5

Now, box B has 7 red and 4 green balls.

P(red ball is picked from B given that green A → B) = 7/11

∴ P(red ball is picked from B) = 3/5 × 7/11 = 21/55

⇒ P(red ball is picked from B) = 16/55 + 21/55 = 37/55

Hence, option (c).

Workspace:

**PE 1 - Probability | Modern Math - Probability**

The letters of the word ‘JUMBLE’ are arranged at random. Find the probability of having exactly two letters in between N and R.

- (a)
1/5

- (b)
1/4

- (c)
1/10

- (d)
1/2

Answer: Option A

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**Explanation** :

There are 6 letters in the word JUMBLE.

Apart from N and R there are 4 more letters.

**Arranging J, E and two letters between them.**

Let us first choose which two letters will be between J and E

This can be done in 4C2 = 6 ways.

These two letters can be arranged in 2! = 2 ways.

J and E can also be arranged in 2! = 2 ways.

⇒ Total ways of arranging J, E and two letters in between them = 6 × 2 × 2 = 24 ways.

Let us call this arrangement X.

**Now we have to arrange X and two remaining letters.**

This can be done in 3! = 6 ways.

⇒ Total ways of arranging the letters of the word JUMBLE such that there are exactly two letters in between J and E = 24 × 6 = 144

Total ways of arranging the letters of the word JUMBLE = 6! = 720

∴ Required probability = 144/720 = 1/5

Hence, option (a).

Workspace:

**PE 1 - Probability | Modern Math - Probability**

400 tickets are numbered as 000, 001, 002, 003,……., 599. If a ticket is drawn at random from them and M is the event that the sum of the digits of the number is 6, then P(M) =

- (a)
1/20

- (b)
9/200

- (c)
1/24

- (d)
None of these

Answer: Option B

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**Explanation** :

Total number of tickets = 600

The possible digits and the number of numbers from 000 to 399 only with those digits, are tabulated below.

Possible digits Number of numbers

0 0 6 → 2

0 1 5 → 6

0 2 4 → 6

0 3 3 → 3

1 1 4 → 3

1 2 3 → 6

2 2 2 → 1

**Total → 27**

∴ Required probability = 27/600 = 9/200

Hence, option (b).

Workspace:

**PE 1 - Probability | Modern Math - Probability**

An unbiased dice is rolled 4 times. Out of 4 numbers that are shown up, the probability that the least is greater than 1 and the greatest is less than 5 is:

- (a)
1/2

- (b)
1/32

- (c)
1/8

- (d)
1/16

Answer: Option D

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**Explanation** :

The numbers that can come up on the dice are 2 or 3 or 5, i.e., 3 possibilities.

⇒ Out of total 6 possibilities there are 3 required outcomes.

∴ Required probability for any 1 toss = 3/6 = ½

⇒ Required probability for 4 tosses = ½ × ½ × ½ × ½ = 1/16.

Hence, option (d).

Workspace:

**PE 1 - Probability | Modern Math - Probability**

Saransh, Akhil and Ronak start a game of dice. They throw the dice by turns, first Saransh, Akhil then Ronak, then once again Saransh, Akhil and Ronak and so on until one of them wins. Saransh is considered to have won if he throws 1 or 2 or 3, Akhil if he throws a prime number and Ronak if he throws 4 or 5 or 6. What is the probability that Saransh wins?

- (a)
3/7

- (b)
4/7

- (c)
1/3

- (d)
Cannot be determined

Answer: Option B

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**Explanation** :

P(S) = P(Saransh getting 1 or 2 or 3) = $\frac{3}{6}$ = $\frac{1}{2}$

P(A) = P(Akhil getting 2 or 3 or 5) = $\frac{3}{6}$ = $\frac{1}{2}$

P(R) = P(Ronak getting 4 or 5 or 6) = $\frac{3}{6}$ = $\frac{1}{2}$

P(Saransh wins) =

P(Saransh gets 1 or 2 or 3 in his first turn) +

P(Saransh gets 1 or 2 or 3 in his second turn) +

P(Saransh gets 1 or 2 or 3 in his third turn) + …

For Saransh to win in his second turn, Akhil and Ronak must not win in their first turn. Same applies for Saransh winning in his further turns.

⇒ P(Saransh wins) = P(S) + P(S’) × P(A’) × P(R’) × P(S) + P(S’) × P(A’) × P(R’) × P(S’) × P(A’) × P(R’) × P(S) + …

= $\frac{1}{2}$ + $\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}$ + $\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}$ +⋯

= $\frac{1}{2}$ + ${\left(\frac{1}{2}\right)}^{4}$ + ${\left(\frac{1}{2}\right)}^{7}$ + ⋯ [Sum of an infinite GP]

= $\frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{1-{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}}$ = $\frac{4}{7}$.

Hence, option (b).

Workspace:

**PE 1 - Probability | Modern Math - Probability**

A letter lock has 4 rings each containing six distinct letters. What is the probability that the code is HIJK in same order?

- (a)
${\left(\frac{1}{{C}_{6}^{26}}\right)}^{4}$

- (b)
1/104

- (c)
${\left(\frac{1}{26}\right)}^{4}$

- (d)
${\left(\frac{1}{{C}_{6}^{26}}\right)}^{4}\times {\left(\frac{1}{6}\right)}^{4}$

Answer: Option C

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**Explanation** :

Probability that H is the 1st letter of the code = $\frac{1}{26}$

Probability that I is the 2nd letter of the code = $\frac{1}{26}$

Probability that J is the 3rd letter of the code = $\frac{1}{26}$

Probability that K is the 4th letter of the code = $\frac{1}{26}$

⇒ Required probability = $\frac{1}{26}$× $\frac{1}{26}$ × $\frac{1}{26}$ × $\frac{1}{26}$ = ${\left(\frac{1}{26}\right)}^{4}$

Hence, option (c).

Workspace:

**PE 1 - Probability | Modern Math - Probability**

A man has five sticks each of length 1 m, 2 m, 3 m, 4 m and 5 m. He chooses three of them randomly and tries to make a triangle. What is the probability that he succeeds?

- (a)
0.2

- (b)
0.7

- (c)
0.3

- (d)
0.1

Answer: Option C

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**Explanation** :

In a triangle, the sum of any two sides is always greater than the third side.

∴ The man can make 3 triangles which satisfy the above criteria i.e., (2, 3, 4), (3, 4, 5) and (2, 4, 5)

⇒ Favourable outcomes = 3

Total outcomes = ^{5}C_{3} = 10

⇒ Required probability = 310 = 0.3

Hence, option (c).

Workspace:

**PE 1 - Probability | Modern Math - Probability**

In a three letter password, what is the probability that all three letters are same?

- (a)
6/26

^{3} - (b)
1/26

^{2} - (c)
1/36

- (d)
None of these

Answer: Option B

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**Explanation** :

Favourable passwords = {AAA, BBB, CCC, ..., ZZZ} i.e., 26 favourable passwords.

Total number of 3 letter passwords that can be formed = 26 × 26× 26 = 26^{3}

∴ Required probability = 26/26^{3} = 1/26^{2}.

Hence, option (b).

Workspace:

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