Mettl Probability Quiz 1

The total number of students = 18
When 1 name was selected from 18 names, the probability that he was of section B
=\frac{6}{3} = \frac{1}{3}
But from the question, there are 6 students from the section B and the age of all 6 are different therefore, the probability of selecting one i.e. youngest student from 6 students will be 1/6

Let, the number of pink balls be p
Probability of choosing a pink ball = \frac{p}{35}
\Rightarrow \frac{3}{7} = \frac{p}{35}
\Rightarrow p = 15
So, remaining number of balls = (35 - 15) = 20
Number of orange balls = \frac{2}{2 + 3}\times 20
= 2 x 4 = 8
Therefore, reqd. Probability \frac{^{8}C_{1}\times ^{15}C_{1}}{^{35}C_{2}}
\frac{8\times 15}{35\times 34\setminus 2} = \frac{24}{119}

Number of ways in which the person can pick three balls out of 18 balls = ¹⁸C₃ = 816
Number of ways of picking 3 balls of same colour

= ⁶C₃ + ⁴C₁ + ⁸C₃ = (20 + 4 + 56) = 80
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 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!} = \frac{16!}{(8\times 8!)}<br /> Ways of correct arrangement = 1<br /> Probability of correct arrangement =<br /> \frac{1}{16!/(8\times 8!)}
\frac{(8\times 8!)}{16!} = \frac{8!}{2\times 15!}

We fix the red face and to its left pink face and bottom face as blue
The number of ways to arrange the other three colors = 3!
Total ways of painting the six colors
First we fix any one color on any one face, let’s say red color.
The number of ways five color can be painted = 5!
Eliminating the repeated possibilities = \frac{5!}{4} = 5\times 3!
We divide by four to eliminate the repeated possibilities as shown in the figure below. These possibilities are counted as different but don’t give us a different arrangement. The arrangement in all four is the same.
Probability (Red, pink and blue share a common corner)
= \frac{3!}{5\times 3!} = \frac{1}{5}

Let the names of children be x, y and z. The probabilities of the three children to finish the race are 1/3, 1/5 and 1/4 respectively. It may be noted that one reaching the finishing point is independent of other reaching. If P(x), P(y) and P(z) denote the probabilities.
The probability of at least one of them reaching the finishing point = 1 – P (none of them finishing the race)
1- (\frac{2}{3})\times (\frac{2}{5})\times (\frac{3}{4}) = \frac{3}{5}

There are 10 even numbers in the group 1-21.
∴ The probability that the first ball is even numbered =\frac{10}{21}
Since the ball is not replaced there are now 20 balls left, of which 9 are even numbered.
∴ The probability that the second ball is even numbered =\frac{9}{20}
∴ Required probability =\frac{10}{21}\times \frac{9}{20} = \frac{9}{42} = \frac{3}{14}

Let E1, E2 be the event of picking a green bulb and white bulb respectively.
Total no. of bulbs in a bag = 3 + 4 + 5 = 12
E_1 = \frac{3}{12} = \frac{1}{4}
E_2 = \frac{5}{12} = \frac{5}{12}
P(E1 or E2) = P(E1) + P(E2)
= \frac{1}{4} + \frac{5}{12} = \frac{2}{3}[/aletx]

The probability of an even number appearing on the first draw is ½
( since there are 25 even numbers in counting of 1 to 50).
The probability of an odd number appearing on the second draw is ½
( since there are 25 odd numbers in counting of 1 to 50).
The probability of a number divisible by 3 appearing on the third draw is 16/50
( since there are 16 numbers that are divisible by 3 while counting from 1 to 50.)
Since all these events have no relation with each other and no dependence either, and the slips are replaced, we can directly multiply the individual probabilities to get the resultant probability.
So, the probability of all the events taking place is
\frac{1}{2}\times \frac{1}{2}\times \frac{16}{50} = \frac{2}{25}

Number of possible combination of 3 persons in which 2 have to be women
= (2 Women out of 5 x 1 Man out of 6) or (3 Women out of 5)
= (⁵C₂ × ⁶C₁ + ⁵C₃)
Total possible outcomes = ¹¹C₃
=\frac{\frac{5!}{2!\times 3!}\times \frac{6!}{5!\times 1!}\times \frac{5!}{3!\times 2!}}{\frac{11!}{3!\times 2!}}
=\frac{70}{11\times 15} = \frac{14}{33}