DevSquare Probability Aptitude Percentages Questions

CORRECT ANSWER IS\frac {23}{56}
HORSE 1: \frac {1}{7} WINNING PROBABILITY
HORSE 2: \frac {1}{8} WINNING PROBABILITY
HORSE 3: \frac {1}{7}WINNING PROBABILITY
ONE OF THESE WIN THE RACE:
=> \frac {1}{7} +\frac {1}{8}+\frac {1}{7}
=> \frac {8}{56} +\frac {7}{56} + \frac {8}{56} (TAKING LCM)
=> \frac {23}{56}

Number of junior students: 1000

Number of Senior Students: 800

60 sibling pair = 2 x 60 = 120 students

The probability of the student being chosen from 800 and 1000 students is 1.

1 student chosen from senior and 1 student chosen from junior

n(s) = 800 x 1000 = 800000

n(E) = 120C2 = 7140

P(E) = n(E)/n(S) = 7140⁄800000

Total no of caps= 5+4+3+2= 14
Probability= ⁵c₂ ⁴c₁ ²c₁ / ¹⁴c₄

80/1001

number of sample solution=36
number of possible outcomes=(1,1),(1,2),(1,4),(1,6),(2,1),(2,3),(2,5),
(3,2)(3,4)(4,1)(4,3)(5,2)

(5,6)(6,1)(6,5)=15
p=15/36

P(A|1)=\frac {3}{7}
P(A|2)=\frac {4}{9}
P(A|3)=\frac {0}{8}
P(A|4)=\frac {2}{9}

Hence event 1,2,3,4 are the partitions of the sample space.

The probability of choosing an apple, by the law of total probability
P(A)=P(A|1)P(1)+P(A|2)P(2)+P(A|3)P(3)+P(A|4)P(4)
=\frac {3}{7}x\frac {1}{4}+\frac {4}{9}x\frac {1}{4}+\frac {0}{8}x\frac {1}{4}+\frac {2}{9}x\frac {1}{4}
=\frac {23}{84}

By Baye\\\'s theorem,
P(2|A)=P(A|2)xP(2)/P(A)
=\frac{\frac {4}{9} \times \frac {1}{4}}{\frac {23}{84}}
=\frac {28}{69}

answer will be 3/4 as total no. of cases of possible moves=8
from this in 2 case their will not be collision(all anti clock or all clock wise)

so,p=(8-2)/8=3/4

No. of events will be 8.
This is because the person can move either inside or outside the triangle.
And the no. of favorable events will be 6. This is because we want
at least 2 person to meet and we have total 3 person, so 2x3 = 6

Probability = 6/8 = 3/4
Ans= 3/4

Let take 100 students
outoff 100 60 are girls
among girls poor girls are =35% 0f 60=21
pbt of selecting poor girl in total strength=21/100

There are 13 spades in a standard deck of cards.

There are four aces in a standard deck of cards.

One of the aces is a spade.

So, 13 + 4 - 1 = 16 spades or aces to choose from.

Since we have a total of 52 cards, the probability of selecting an ace
or a spade is 16 / 52.

ans = 1/3
becoz ther is only one chance to get tail on other side of coin,
amnog three coins,according to the
question