AMCAT Probability Quiz-1

Answer : 0.46
Solution :
The probability of selection of Prashant is 0.3
The probability of not selection of Prashant is 0.7

The probability of selection of Anmol is 0.4
The probability of not selection of Prashant is 0.6

The probability that any one of them is selected = 0.3 x 0.6 + 0.7 x 0.4 = 0.46

Answer : 0.74
Explanation :
probability of not winning for the Indian team = 1- 0.26 = 0.74

Answer : \frac{4}{5}

Explanation :
Non defective items = 20 - 4 = 16
Probability of non defective items = \frac{16}{20} = \frac{4}{5}

Answer : 1/2
Explanation:
All Events are Mutually Exclusive.
Only one wins Race means Either Tesla car or Porsche car.
Probability that only one of them win Race = 1/10+4/10 = 5/10 = ½

Answer : \frac{1}{4}
Explanation :
Two digit total possible numbers using digits 3, 9, 1, and 8 = 4 x 4 = 16
For a number to be divisible by 2 here unit digits are only 8 possible
Then the total number of possible combinations for divisible by 2 = 4 x 1 = 4
Required Probability = \frac{4}{16} = \frac{1}{4}

Answer : \frac{3}{5}
Explanation :
Let the names of racers be L, S and D.
The probabilities of the three racers 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 the other reaching.
If P(L), P(S) and P(D) denote the probabilities.
The probability of at least one of them reaching the finishing point = 1 - P (none of them finishing the race)

=1 - \left ( \frac{2}{3} \right )\left ( \frac{4}{5} \right )\left ( \frac{3}{4} \right ) = \frac{3}{5}

Answer : = \frac{5}{16}
Explanation :
When 4 fair coins are tossed simultaneously, the total number of outcomes is 2^4 = 16
At least 3 heads imply that one can get either 3 heads or 4 heads.
One can get 3 heads in ^4C_3 = 4 ways and can get 4 heads in ^4C_4 = 1 ways.
∴ Total number of favorable outcomes = 4 + 1 = 5
∴ The reqd. probability = \frac{5}{16}

Answer : \frac{1}{18}
Explanation :
The total number of students = 18
When 1 name was selected from 18 names, the probability that he was of section B
=6/18= 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

Probability =\frac{1}{6} \times \frac{1}{3} = \frac{1}{18}

Answer : \frac{8!}{2\times 15!}
Explanation :
total pieces in chess=32
So in 2 rows, 16 pieces can be placed
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

Ways of correct arrangement = 1
Probability of correct arrangement = \frac{1}{\frac{16!}{8\times 8!}}
\frac{(8\times 8!)}{16!}= \frac{8!}{2\times 15!}

Answer : \frac{4}{17}
Explanation :
Let us assume that all balls are unique.
There are a total of 18 balls.
Total ways = 3 balls can be chosen in ^{18}\textrm{C}_{3} ways
= \frac{18!}{3! \times 15!} = \frac{18 \times 17 \times 16}{3 \times 2 \times 1} = 816
Favorable ways = 1 tennis ball, 1 season ball, and 1 dues Ball = 4 × 6 × 8 = 192
Probability = \frac{192}{816} = \frac{4}{17}