Probability For IBPS RRB 2024 Exam: Probability is a short topic in the Quantitative Aptitude section of the IBPS RRB 2024 exam. If candidates Understand the formulas and principles of probability then you can solve every question for the probability section. This article provides comprehensive details for Probability sections for the IBPS RRB 2024 exam along with the concepts, formulas and some questions with proper solutions.
Probability is the meaning of the likelihood of an event occurring. Probability is an occurring event. It ranges from 0 to 1, where 0 indicates that the event cannot happen, and 1 indicates certainty that the event will happen. The basic formula for probability is:
i.e., P(A) = n(A)/n(S)
where,
Question 1: A bag contains balls of three different colours such that the number of green and blue balls are in the ratio of 3:4 respectively while the number of red balls is 20. If probability of getting a green ball is 0.25, then find total number of balls in the bag.
A) 48
B) 34
C) 41
D) 27
E) None of these
Question 2: A bag contains ‘x’ green and ‘x + 1’ red balls. If probability of drawing a red and a green ball from the bag simultaneously is 5/9, then find number of red balls in the bag.
A) 7
B) 9
C) 4
D) 5
E) None of these
Question 3: A bag containing 24 balls has red, blue and green balls in the ratio of 2:3:1, respectively. Two balls are drawn at random from the bag, then what is the probability that both drawn balls are of red colour.
A) 3/52
B) 7/69
C) 11/63
D) 4/53
E) None of these
Question 4: Two dices are thrown together. What is the probability of getting a sum between 8 and 12?
A) 7/36
B) 1/4
C) 5/36
D) 1/6
E) None of these
Question 5: Two dices are thrown together. What is the probability of getting a sum of either 6 or 8?
A) 7/18
B) 11/36
C) 1/4
D) 3/5
E) 5/18
Question 6: A box contains 12 articles out of which 4 are defective. If 6 articles are drawn randomly, then find the probability that out of 6 articles drawn only 1 article is defective.
A) 6/55
B) 4/231
C) 5/44
D) 7/22
E) 8/33
Question 7: A word consist of 7 letters i.e. 4 consonant and 3 vowels. 3 letters are selected at random. Find the probability that more than one consonant is selected.
A) 22/35
B) 11/34
C) 13/35
D) 12/35
E) None of these
Question 8: A basket contains 3 apples and 7 mangoes. Two fruits are drawn at random. Find the probability that both the fruits are same.
A) 8/15
B) 2/9
C) 1/12
D) 7/15
E) 4/15
Question 9: Find the number of arrangements of the word “POSSIBILITIES” so that all the vowels always occur together.
A) 201600
B) 403200
C) 100800
D) 302400
E) None of these
Question 10: Number of green balls in a bag is 6 more than number of red balls which is 2 less than number of blue balls in it. If total number of balls in the bag is 50, then probability of drawing three different colour balls when three balls are drawn from the bag will be:
A) 8/35
B) 4/35
C) 16/35
D) 1/4
E) None of these
प्रश्न 1: एक bag में तीन विभिन्न रंग के balls है जहाँ green और blue balls की संख्या का अनुपात क्रमशः 3:4 है जबकि red balls की संख्या 20 है|यदि एक green ball को प्राप्त करने की प्रायिकता 0.25 है, तो bag में balls की कुल संख्या ज्ञात करें|
A) 48
B) 34
C) 41
D) 27
E) इनमे से कोई नहीं
प्रश्न 2: एक बैग में ‘x’ green और ‘x + 1’ red balls हैं। यदि बैग से एक red और green ball एक साथ निकालने की प्रायिकता 5/9 है, तो बैग में red balls की संख्या ज्ञात कीजिए।
A) 7
B) 9
C) 4
D) 5
E) इनमें से कोई नहीं
प्रश्न 3: 24 balls वाले एक बैग में red, blue और green balls का अनुपात क्रमशः 2:3:1 है|बैग से दो balls का चुनाव किया जाता है, तो इस बात की कितनी प्रायिकता होगी कि निकाले गए दोनों balls red रंग के होंगे?
A) 3/52
B) 7/69
C) 11/63
D) 4/53
E) इनमे से कोई नहीं
प्रश्न 4: दो पासों को साथ में फेंका जाता है|8 और 12 के बीच के योग को प्राप्त करने की प्रायिकता कितनी है?
A) 7/36
B) 1/4
C) 5/36
D) 1/6
E) इनमे से कोई भी नहीं
प्रश्न 5: दो पासे को साथ में फेंका जाता है|या तो 6 या 8 का योग ज्ञात करने की प्रायिकता कितनी है?
A) 7/18
B) 11/36
C) 1/4
D) 3/5
E) 5/18
प्रश्न 6: एक box में 12 वस्तुएं हैं जिनमें से 4 ख़राब होते हैं। यदि 6 वस्तुओं को यदृछया निकाला जाता हैं, तो इस बात की प्रायिकता ज्ञात करें कि 6 वस्तुओं में से केवल 1 वस्तु ख़राब है।
A) 6/55
B) 4/231
C) 5/44
D) 7/22
E) 8/33
प्रश्न 7: एक शब्द में 7 अक्षर अर्थात 4 consonant और 3 vowels होते हैं। 3 अक्षर यादृच्छिक पर चुने गए हैं। एक से अधिक consonant चुने जाने की प्रायिकता ज्ञात कीजिए।
A) 22/35
B) 11/34
C) 13/35
D) 12/35
E) इनमें से कोई नहीं
प्रश्न 8: एक टोकरी में 3 सेब और 7 आम हैं। यादृच्छिक रूप से दो फल निकाले जाते हैं। दोनों फलों के समान होने की प्रायिकता ज्ञात कीजिए।
A) 8/15
B) 2/9
C) 1/12
D) 7/15
E) 4/15
प्रश्न 9: शब्द “POSSIBILITIES” की व्यवस्थाओं की संख्या ज्ञात करें ताकि सभी vowels हमेशा एक साथ हों।
A) 201600
B) 403200
C) 100800
D) 302400
E) इनमें से कोई नहीं
प्रश्न 10: एक बैग में green balls की संख्या red balls की संख्या से 6 अधिक है जो उसमें blue balls की संख्या से 2 कम है। यदि बैग में balls की कुल संख्या 50 है, तो बैग से तीन balls निकालने पर तीन अलग-अलग रंग की balls निकालने की प्रायिकता कितनी होगी:
A) 8/35
B) 4/35
C) 16/35
D) 1/4
E) इनमें से कोई नहीं
ANSWER KEYS and SOLUTIONS:
1) – A) | 2) – D) | 3) – B) | 4) – B) | 5) – E) | 6) – E) |
7) – A) | 8) – A) | 9) – A) | 10) – A) |
Solution 1: A)
Let number of green and blue balls in the bag is ‘3x’ and ‘4x’ respectively.
Total number of balls in the bag = 3x + 4x + 20 = 7x + 20
According to question;
3x/(7x + 20) = 1/4
Or, 12x = 7x + 20
Or, 5x = 20
Or, x = 4
Total number of balls in the bag = 7 × 4 + 20 = 48
Hence, option a.
Solution 2: D)
Total number of balls in the bag = x + x + 1 = 2x + 1
So, {(xC1 × (x + 1)C1)/(2x + 1)C2} = 5/9
Or, {2x(x + 1)/(2x + 1)2x} = 5/9
Or, 9x + 9 = 10x + 5
Or, x = 4
So, number of red balls = x + 1 = 4 + 1 = 5
Hence, option d.
Solution 3: B)
Number of red balls in bag = 2/6 × 24 = 8
Required probability = 8C2/ 24C2= 7/69
Hence, option b.
Solution 4: B)
Total number of outcomes = 6 × 6 = 36
Favourable outcomes are (3,6), (4,5), (5,4), (6,3), (4,6), (5,5), (6,4), (6,5), (5,6)
Total number of favourable outcomes = 9
Desired probability = 9/36 = 1/4
Hence, option b.
Solution 5: E)
Total number of outcomes = 6 × 6 = 36
Favourable outcomes = (3,3), (4,2), (5,1), (2,4), (1,5), (6,2), (2,6), (5,3), (3,5), (4,4)
Total number of favourable outcomes = 10
Desired probability = 10/36 = 5/18
Hence, option e.
Solution 6: E)
Since, 6 articles are drawn randomly
Therefore, total number of ways of selecting 6 articles = 12C6 = 12!/{6! × (12 – 6)!} = 924
Number of ways of selecting 5 non-defective articles out of 8 = 8C5 = 56 ways
Number of ways of selecting 1 defective article out of 4 defective articles = 4C1 = 4
Required probability = Number of ways of getting desired outcome/Total number of outcomes = (56 × 4)/924 = 8/33
Hence, option e.
Solution 7: A)
Number of ways of selecting 3 letters out of 7 = 7C3 = 35 ways
Number of ways of selecting more than 1 consonant = (2 consonants and 1 vowel + 3 consonants)
Required number of ways = 4C2 × 3C1 + 4C3 = 6 × 3 + 4 = 22 ways
Required probability = 22/35
Hence, option a.
Solution 8: A)
Total number of fruits in the basket = 3 + 7 = 10
Number of ways of withdrawing 2 fruits = 10C2 = (10 × 9)/2 = 45
Number of ways of withdrawing same fruits = 3C2 + 7C2 = 3 + 21 = 24
Required probability = 24/45 = 8/15
Hence, option a.
Solution 9: A)
Desired number of ways = (8! × 6!)/(4! × 3!) = 201600
Hence, option a.
Solution 10: A)
Let number of red balls be ‘x’
So, number of green ball = ‘x + 6’
Number of blue ball = x + 2
So, x + x + 6 + x + 2 = 50
Or, 3x = 42
Or, x = 14
So, number of red balls = 14
Number of green balls = 14 + 6 = 20
Number of blue balls = 14 + 2 = 16
Desired probability = {16C1 × 20C1 × 14C1}/50C3 = (16 × 20 × 14 × 6)/(50 × 49 × 48) = 8/35
Hence, option a.
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Probability is the measure of the likelihood that an event will occur. It ranges from 0 to 1.
An event is a subset of the sample space. It consists of one or more outcomes.
The Formula for calculating probability is P(E)= Number of favorable outcomes/Total number of possible outcomes
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