NO Search

Q:

A basket contains 10 apples and 20 oranges out of which 3 apples and 5 oranges are defective. If we choose two fruits at random, what is the probability that either both are oranges or both are non defective?

A) 136/345	B) 17/87
C) 316/435	D) 158/435

Answer & Explanation Answer: C) 316/435

Explanation:

$n (s) = {}^{30}C_{2}$

Let A be the event of getting two oranges and

B be the event of getting two non-defective fruits.

and $(A \cap B)$ be the event of getting two non-defective oranges

$∴ P (A) = \frac{{}^{20}C_{2}}{{}^{30}C_{2}}, P (B) = \frac{{}^{22}C_{2}}{{}^{30}C_{2}} a n d P (A \cap B) = \frac{{}^{15}C_{2}}{{}^{30}C_{2}}$

$∴ P (A \cup B) = P (A) + P (B) - P (A \cap B)$

= $\frac{{}^{20}C_{2}}{{}^{30}C_{2}} + \frac{{}^{22}C_{2}}{{}^{30}C_{2}} - \frac{{}^{15}C_{2}}{{}^{30}C_{2}} = \frac{316}{435}$

Report Error

View Answer Report Error Discuss

Filed Under: Probability

Q:

In a single throw of two dice , find the probability that neither a doublet nor a total of 8 will appear.

A) 7/15	B) 5/18
C) 13/18	D) 3/16

Answer & Explanation Answer: B) 5/18

Explanation:

n(S) = 36

A = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}

B = { (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) }

$n (A) = 6, n (B) = 5, n (A \cap B) = 1$

$∴$ Required probability = $P (A \cup B)$

= $P (A) + P (B) - P (A \cap B)$

= $\frac{6}{36} + \frac{5}{36} - \frac{1}{36}$ = $\frac{5}{18}$

Report Error

View Answer Report Error Discuss

Filed Under: Probability

Q:

The probability of occurance of two two events A and B are 1/4 and 1/2 respectively. The probability of their simultaneous occurrance is 7/50. Find the probability that neither A nor B occurs.

A) 25/99	B) 39/100
C) 61/100	D) 17/100

Answer & Explanation Answer: B) 39/100

Explanation:

P ( neither A nor B) = $P (A a n d B)$

= $P (A \cap B)$ = $P (A \cup B)$ = $1 - P (A \cup B)$

= $1 - \frac{61}{100} = \frac{39}{100}$

Report Error

View Answer Report Error Discuss

Filed Under: Probability

Q:

A box contains 5 detective and 15 non-detective bulbs. Two bulbs are chosen at random. Find the probability that both the bulbs are non-defective.

A) 5/19	B) 3/20
C) 21/38	D) None of these

Answer & Explanation Answer: C) 21/38

Explanation:

n(S) = ${}^{20}C_{2}$ = 190

n(E) = ${}^{15}C_{2}$ = 105

Therefore, P(E) = 105/190 = 21/38

Report Error

View Answer Report Error Discuss

Filed Under: Probability

Q:

One card is drawn from a pack of 52 cards , each of the 52 cards being equally likely to be drawn. Find the probability that the card drawn is neither a spade nor a king.

A) 0	B) 9/13
C) 1/2	D) 4/13

Answer & Explanation Answer: B) 9/13

Explanation:

There are 13 spades ( including one king). Besides there are 3 more kings in remaining 3 suits

Thus n(E) = 13 + 3 = 16

Hence $n (\bar{E}) = 52 - 16 = 36$

Therefore, $P (E) = \frac{36}{52} = \frac{9}{13}$

Report Error

View Answer Report Error Discuss

Filed Under: Probability

Q:

Find the number of ways in which 21 balls can be distributed among 3 persons such that each person does not receive less than 5 balls.

A) 28	B) 14
C) 21	D) 7

Answer & Explanation Answer: A) 28

Explanation:

Let x, y, z be the number of balls received by the three persons, then

$x \geq 5, y \geq 5, z \geq 5 a n d x + y + z = 21$

Let $u \geq 0, v \geq 0, w \geq 0 t h e n$

$∴$ x + y + z =21

$\Rightarrow$ u + 5 + v + 5 + w + 5 = 21

$\Rightarrow$ u + v + w = 6

$∴$ Total number of solutions = ${}^{6 + 3 - 1}C_{3 - 1} = {}^{8}C_{2} = 28$

Report Error

View Answer Report Error Discuss

Filed Under: Permutations and Combinations

Q:

In a Plane there are 37 straight lines, of which 13 pass through the point A and 11 pass through the point B. Besides, no three lines pass through one point, no lines passes through both points A and B , and no two are parallel. Find the number of points of intersection of the straight lines.

A) 525	B) 535
C) 545	D) 555

Answer & Explanation Answer: B) 535

Explanation:

The number of points of intersection of 37 lines is $C_{2}^{37}$ . But 13 straight lines out of the given 37 straight lines pass through the same point A.

Therefore instead of getting $C_{2}^{13}$ points, we get only one point A. Similarly 11 straight lines out of the given 37 straight lines intersect at point B. Therefore instead of getting $C_{2}^{11}$ points, we get only one point B.

Hence the number of intersection points of the lines is $C_{2}^{37} - C_{2}^{13} - C_{2}^{11} + 2$ = 535

Report Error

View Answer Report Error Discuss

Filed Under: Permutations and Combinations
Exam Prep: CAT , Bank Exams , AIEEE
Job Role: Bank PO , Bank Clerk

Q:

How many arrangements can be made out of the letters of the word COMMITTEE, taken all at a time, such that the four vowels do not come together?

A) 216	B) 45360
C) 1260	D) 43200

Answer & Explanation Answer: D) 43200

Explanation:

There are total 9 letters in the word COMMITTEE in which there are 2M's, 2T's, 2E's.

The number of ways in which 9 letters can be arranged = $\frac{9!}{2! \times 2! \times 2!}$ = 45360

There are 4 vowels O,I,E,E in the given word. If the four vowels always come together, taking them as one letter we have to arrange 5 + 1 = 6 letters which include 2Ms and 2Ts and this be done in $\frac{6!}{2! \times 2!}$ = 180 ways.

In which of 180 ways, the 4 vowels O,I,E,E remaining together can be arranged in $\frac{4!}{2!}$ = 12 ways.

The number of ways in which the four vowels always come together = 180 x 12 = 2160.

Hence, the required number of ways in which the four vowels do not come together = 45360 - 2160 = 43200

Report Error

View Answer Report Error Discuss

Filed Under: Permutations and Combinations

Searching for "NO"

Quick Links