HackerRank Probability Quiz -1

Probability of getting a English book = 6/(5+6) = 6/11

The total number of students = 18

When 1 name was selected from 18 names, the probability that he was of House Indigo = \frac{6}{18} = \frac{1}{3}

But from the question, there are 6 students from the House Indigo and the ages of all 6 are different.

Therefore, the probability of selecting one i.e. youngest student from 6 students will be ⅙

Let, the number of Green balls be p

Probability of choosing a Green ball = \frac{p}{35}

=>\frac{3}{7} = \frac{p}{35}

=> p = 15

So, remaining number of balls = (35 – 15) = 20

Number of Blue balls = \frac{2}{(2+3)} \times 20 = 8

Therefore, reqd. probability =\frac{_{1}C^{8}\textrm{}\times _{1}C^{15}\textrm{}}{_{2}C^{35}\textrm{}}

= \frac{24}{119}

Number of ways in which Rajesh can pick three balls out of 18 balls = _{3}C^{18}\textrm{} = 816

Number of ways of picking 3 balls of same colour = _{3}C^{6}\textrm{}+ _{1}C^{4}\textrm{} + _{3}C^{8}\textrm{}\frac{80}{816} = \frac{5}{51}

Probability of picking three balls of same color =\frac{80}{816} = \frac{5}{51}

Required probability = 1 – probability of picking three balls of same colour = 1 – \frac{5}{51} = \frac{46}{51}

Total sample space = 6 x 6 = n(S) =36

Event of getting numbers whose product is an even number = n(E) = {(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3, 4), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

n (E) = 27

Probability n(E)/n(S) = 27/36=¾

Sample space = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}

Let us assume the event = {TTT, TTH, THT, HTT, THH, HTH, HHT}

= n(E)/n(S) = ⅞

Total boxes = 16

Total pieces = 16

Similar pieces = 8 pawns, 2 bishops, 2 rooks, 2 knights

Total ways of arranging these 16 pieces in 16 boxes

= \frac{16!}{8! 2! 2! 2!}

Ways of correct arrangement = 1

Probability of correct arrangement = \frac{1}{\frac{16!}{8 \times 8!}}

= \frac{(8 \times 8!)}{16!}

We fix the violet face and to its left blue face and bottom face as indigo.

The number of ways to arrange the other three colors = 3!

First we fix any one color on any one face, let’s say violet color.

The number of ways five colors can be painted = 5!

Eliminating the repeated possibilities = \frac{5!}{4} = 5 \times 3!

These possibilities are counted as different but don’t give us a different arrangement. The arrangement in all four is the same.

Probability (violet, blue and indigo share a common corner)=

\frac{3!}{5 \times 3!} = ⅕

Total sample space = 16.

Favorable outcomes = 1 as the required combination occurs only in one way.

Probability = 1/16

Total sample space = n(S)

=6 * 6

= 36

We know,

Event of getting favourable outcomes = n(E)

= {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}

Total probability = n(E)/n(S) = 5/36