Quantitative Aptitude: Probability Questions – Set 22
- Two dice are rolled randomly. Find the probability to get sum is 10.
2/5142/91/125/8Option D
Required = (6 , 4 ) , (4 , 6) , ( 5, 5) = 3
total = 6 * 6 = 36
probability = 3/36 = 1/12 - Two dices are rolled out together, then what is the probability of getting a number of one dice greater than the number of other dice ?
2/31/63/81/85/6Option E
Non-favorable events = (1 , 1) , (2 , 2) , (3 , 3) , (4 , 4) , (5 , 5) , (6,6) = 6
total = 6 * 6 = 36
probability of non-favorable events = 6/36 = 1/6
probability of favorable events = 1 – 1/6 = 5/6 - If 2 cards is drawn at randomly from 52 cards, then find the probability of getting both are red cards.
13/2826/10528/10825/102none of theseOption D
Required = 26C2 = 26 * 25/2 = 13 * 25
total = 52C2 = 52 * 51 / 2 = 26 * 51
probability = 13 * 25 / 26 * 51 = 25 /102 - In how many ways, we can arrange the letters of the word ‘LIGHT’ ?
8428140160120Option E
Ways = 5! = 5 * 4 * 3 * 2 * 1 = 120 - In how many different ways, we can arrange the letters of the word ‘MOUSE’ , so that the middle position is always occupied by ‘S’ ?
2524484268Option B
Ways = 4! = 4 * 3 * 2 * 1 = 24 - How many ways 6 books can be selected from 14 different books , if two particular books are always selected ?
495480231384520Option A
Available objects = 14 – 2 = 12
Ways = 12C4 = 12 * 11 * 10 * 9/4 * 3 * 2 * 1 = 495 - A bag contains 2 white balls, 3 pink balls and 2 black balls. 2 balls are drawn randomly. What is the probability that there is no black balls ?
8/2110/211/83/56/11Option B
Required = 5C2 = 5*4/2 = 10
total = 7C2 = 7 * 6 / 2 = 21
probability = 10/21 - When two coins are tossed simultaneously, then find the probability of getting at least one tail.
533/41/42/5Option C
Required = ( 1 head , 1 tail ) , ( 2 tails) = (2C1) + (2C2) = 2 + 1 = 3
total = 2^2 = 4
probability = 3/4 - A bag contains 4 red balls, 2 blue balls and 2 green balls . If two balls are drawn randomly from the bag , then find the probability of getting both balls of different color.
2/75/85/748Option C
probability = 4C1 * 2C1 / 8C2 + 4C1 * 2C1 / 8C2 + 2C1 * 2C1 / 8C2
(4 * 2 * 2)/ 8 * 7 + (4 * 2 * 2)/ 8 * 7 + (2 * 2 * 2) / 8 * 7 = 2/7 + 2/7 + 1/7 = 5/7 - A group of students sitting around a rectangular table, find probability of 2 specified students sitting together.
3/82/78/73/86/11Option B
Favorable cases = 6! * 2!
total cases = 7!
probability = 6! * 2! /7! = 2/7
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