41. A box contains, 5 Sony, 6 Samsung and 4 SanDisk pen drives. 3 pen drives are drawn at random. What is the probability that they are from different company?
एक बॉक्स में 5 सोनी, 6 सैमसंग और 4 सैनडिस्क पेन ड्राइव हैं। यादृच्छिक रूप से 3 पेन ड्राइव निकाले जाते हैं। क्या संभावना है कि वे अलग-अलग कंपनी से हैं?A. 24/91
B. 27/85
C. 33/67
D. 35/78
E. None of these
Option “A” is correct.

Total probability n(S) = 15C3

Required probability n(E) = 5C1 and 6C1 and 4C1P(E) = n(E)/n(S)P(E) = 5C1 and 6C1 and 4C1 / 15C3P(E) = 24/91
42. A basket contains x apple, 4 orange and 3 banana. One fruit is taken out randomly and the probability of getting a orange is 2/5, and find the value of x?
एक टोकरी में x सेब, 4 संतरे और 3 केले हैं। एक फल यादृच्छया निकाला जाता है और एक संतरा प्राप्त होने की प्रायिकता 2/5 है, फिर x का मान ज्ञात कीजिए?A. 2
B. 3
C. 1
D. 4
E. None of these
Option “B” is correct.

4C1/(7 + x)C1 = 2/520 = 14 + 2x2x = 6x = 3
43. A bag contains 45 balls marked 1 to 45. If one ball is drawn at random, then what is the probability that is marked with a number divisible by 4?
एक बैग में 1 से 45 तक चिह्नित 45 गेंदें हैं। यदि एक गेंद यादृच्छिक रूप से निकाली जाती है, तो गेंद के 4 से विभाज्य संख्या के साथ चिह्नित होने की क्या संभावना है?A. 4/15
B. 3/5
C. 2/5
D. 8/45
E. None of these
Option “E” is correct.

Number of marked balls divisible by 4 = 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44= 11

Required probability = 11/45
44. A bag contains 30 balls numbered from 1 to 30. Two balls are selected one by one with replacement at random. What is the probability that the first ball is a prime number and the second ball is a perfect square?
एक बैग में 1 से 30 तक चिह्नित 30 गेंदें हैं। यादृच्छिक रूप से प्रतिस्थापन के साथ दो गेंदों को एक-एक करके चुना जाता है। पहली गेंद एक अभाज्य संख्या है और दूसरी गेंद एक पूर्ण वर्ग है, इसकी क्या प्रायिकता है?A. 1/3
B. 1/6
C. 1/9
D. 1/18
E. 1/24
Option “D” is correct.

Total balls = 30 Total prime numbers between 1 to 30 = 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 = 10

Total perfect square = 1, 4, 9, 16, 25 = 5

Probability of selecting first ball = 10/30 = 1/3

Probability of selecting second ball = 5/30 = 1/6

Therefore, probability = (1/3) * (1/6) = 1/18
45. A bag contains 8 balls out which x red balls and remaining are yellow balls. If two balls are drawn at random and the probability that both are red balls is 5/14, then find the number of yellow balls in the bag?
एक बैग में 8 गेंदें हैं जिनमें से x लाल गेंदें और शेष पीली गेंदें हैं। यदि दो गेंदों को यादृच्छिक रूप से निकाला जाता है और दोनों के लाल होने की प्रायिकता 5/14 है, तो बैग में पीली गेंदों की संख्या ज्ञात कीजिए?A. 1
B. 3
C. 2
D. 4
E. 6
Option “B” is correct.Number of red balls=xNumber of yellow balls=(8 – x)5/14=xC2/8C2   5/14=x * (x – 1)/5620=x2 – x

x2 – x – 20=0

x2 – 5x + 4x – 20=0

x(x – 5) + 4(x – 5)=0

x=-4, 5(negative value neglected)

x=5

number of yellow balls=8 – 5=3
46. How many 3 letter words with or without meaning can be formed, out of the letters of the word, ‘BIFURCATE’, if repetition of letters is not allowed?
यदि अक्षरों की पुनरावृत्ति की अनुमति नहीं है, तो शब्द ‘BIFURCATE’ के अक्षरों से, अर्थ सहित या बिना अर्थ के कितने 3 अक्षर वाले शब्द बन सकते हैं?A. 84
B. 540
C. 504
D. 512
E. None of these
Option “C” is correct.‘BIFURCATE’ contains 9 different letters.Required number of words = 9P3= > (9 * 8 * 7) = 504
47. In a group of 4 boys and 3 girls, three children are to be selected. In how many different ways can they be selected such that at least one boy should be there?
4 लड़कों और 3 लड़कियों के समूह में, तीन बच्चों का चयन किया जाना है। उन्हें कितने अलग – अलग तरीकों से चुना जा सकता है जैसे कि कम से कम एक लड़का होना चाहिए?A. 60
B. 35
C. 42
D. 34
E. 38
Option “D” is correct.At least one boy = Total ways – ways of no boys= 7C3 – 3C3= (7 × 6 × 5)/(3 × 2 × 1) – 1= 35 – 1= 34
48. In how many different ways can the letters of the word ‘COMBINATION’ be arranged so that the vowels always come together?
शब्द ‘COMBINATION’ के अक्षरों को कितने अलग-अलग तरीकों से व्यवस्थित किया जा सकता है ताकि स्वर हमेशा एक साथ आएं?A. 26800
B. 56700
C. 75600
D. 13600
E. None of these
Option “C” is correct.COMBINATION = CMBNTN(OIAIO)Number of ways of arranging these letters = 7! / 2! = 2520Number of ways of arranging OIAIO = 5! /(2! * 2!) = 30Required number of words = 2520 * 30 = 75600
49. How many 7 letter words with or without meaning can be formed out of the letters of the word, “ENCAPSULATION”?
“ENCAPSULATION” शब्द के अक्षरों से अर्थ सहित या बिना अर्थ के कितने 7 अक्षर वाले शब्द बनाए जा सकते हैं?A. 78280
B. 6282620
C. 8648640
D. 56600
E. None of these
Option “C” is correct.“ENCAPSULATION” contains 13 different letters.Required number of words = 13p7= > (13 * 12 * 11 * 10 * 9 * 8 * 7) = 8648640
50. In how many ways 6 Teachers, 7 Doctors and 8 Engineers be seated in a row in the conference hall, so that all person of same profession sits together?
सम्मेलन हॉल में 6 शिक्षक, 7 डॉक्टर और 8 इंजीनियरों को कितने तरीकों से एक पंक्ति में बैठाया जाए, ताकि एक ही पेशे के सभी व्यक्ति एक साथ बैठें?A. 6!7!8!1!
B. 6!7!8!2!
C. 6!7!8!3!
D. 6!7!8!4!
E. None of these
Option “C” is correct.6 Teachers can be seated together in 6! ways.Similarly for Doctors and Engineers in 7! and 8! respectively.Required number of ways = 6! 7! 8! 3!
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