Mahindra Comviva Probability Problems with Solutions Quiz -1

Let S be the sample space.

Then, n(S) = number of ways of drawing 3 balls out of 15
= 15C3
= (15 x 14 x 13)
(3 x 2 x 1)
= 455.
Let E = event of getting all the 3 red balls.

n(E) = 5C3 = 5C2 = (5 x 4) = 10.
(2 x 1)
P(E) = n(E) = 10 = 2 .
n(S) 455 91

Please remember that Maximum portability is 1.

So we can get total probability of non defective bulbs and subtract it form 1 to get total probability of defective bulbs.

So here we go,
Total cases of non defective bulbs
16C2=16∗152∗1=120total cases = 20C2=20∗192∗1=190probability = 120190=1219P of at least one defective = 1−1219=719

Probability of English P(E) = 30 / 100 = 3 / 10,
Probability of Hindi P(H) = 20 / 100 = 2 / 10 and
Probability of Both Hindi and English P(E and H) = 10 / 100 = 1 / 10
P (E or H) = P (E) + P (H) - P (E and H)
= (3 / 10) + (2 / 10) - (1 / 10)
= 4 / 10
= 2 / 5

Given that the ball is red, there are 6 red balls, so the probability that it's the smallest of those is 1/6.

P(A/B) = P(A ∪ B)/P(B)

P(A ∪ B) = P(A)P(B) = 0:2/0.4 =0.5

P = Number of desired outcomes / number of possible outcomes

Number of desired outcomes:
To be divisible by 4, the number has to end with the last two digits forming a number divisible by 4. Given the numbers in the question this leaves us with:
12
24
32
52
For each of them I now have to put the remaining numbers in front: 3x2x1 --> 6 combinations for each number set = 6x4 (we have 4 sets from above) = 24

Number of possible outcomes:
The total number of possibilities to form a 5 digit number from the question= 5x4x3x2x1=120

P = 24/120 = 1/5

There are six numbers on a dice: 1, 2, 3, 4, 5, and 6. Hence total number of outcomes possible on throw of a dice once are six (any one of the six numbers can show up). There are three even numbers on a dice: 2, 4 and 6. Hence total number of outcomes favorable to us are three (since any one of these three even numbers can show up). Hence by applying the formula for theoretical probability, we get,
`P(""outcome"") = ""number of favorable outcomes""/""total number of possible outcomes""`
`P(""even number"") = 3/6 = 1/2`
Hence the probability of getting an even number when throwing a dice once is 1/2. Similarly, the probability of getting an odd number on a single throw of a dice is 1/2

Total outcomes = 12
he wants to apple or banana, thus no. of apple and banana together = 4+3=7
probability= favourable outcomes/total outcomes
= 7/12

Red ball from box A = 3
red ball from box B = 5

red from A and B = 3/7 * 5/7 = 15/49