- QUESTION
A glass jar contains 1 red, 3 green, 2 blue and 4 yellow marbles. If a single marble is chosen at random from the jar
(A)Probability of getting a yellow marble is 1/5
(B)Probability of getting a green marble is 3/10
(C)Probability of getting either a yellow or a green marble is 7/10
(D)Probability of getting either a red or a yellow marble is 3/10
Choose the correct answer from the options given below:
A)(A) and (D) only
B)(B) and (C) only
C)(A) and (C) only
D) (B) and (D) only
Explanation
(B)
=Let’s calculate the probabilities:
Total number of marbles = 1 (red) + 3 (green) + 2 (blue) + 4 (yellow) = 10.
(A) Probability of getting a yellow marble = Number of yellow marbles / Total number of marbles = 4/10 = 2/5. So, option (A) is incorrect.
(B) Probability of getting a green marble = Number of green marbles / Total number of marbles = 3/10. So, option (B) is correct.
(C) Probability of getting either a yellow or a green marble = Probability of yellow marble + Probability of green marble = (4/10) + (3/10) = 7/10. So, option (C) is correct.
(D) Probability of getting either a red or a yellow marble = Probability of red marble + Probability of yellow marble = (1/10) + (4/10) = 1/2. So, option (D) is incorrect.
Therefore, the correct answer is (B) and (C) only.
QUESTION-2
Find the number of permutations that can be made from the letters of the word OMEGA, if the vowels occupy odd places.
A)12 WAYS
B)6 WAYS
C)13 WAYS
D)17 WAYS
Explanation
(A)
= 12 WAYS
OMEGA
Here, no. Of vowels=3(o,e,a)
No. Of consonants =2(m,g)
Now vowels are to occupy odd places
There are 3 odd places of in which the 3 vowels can be arranged in 3! ways=3x2x1=6
In the remaining 2even places 2consonants can be arranged in 2! ways=2×1=2 ways.
Thus required no. of words formed
= 6×2
=12
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