A. Mesiobuccal
B. Distobuccal
C. Mesio lingual
D. Distal
A. Mesiobuccal
B. Distobuccal
C. Mesio lingual
D. Distal
A. Maxillary lateral incisor
B. Maxillary first premolar
C. Maxillary first molar
D. All of the above
A. Oval
B. Trapezoidal
C. Triangular
D. Rhomboidal
A. 1/26
B. 3/52
C. 15/26
D. 11/26
Explanation:
Let E1 be the event of drawing a red card.
Let E2 be the event of drawing a king .
P(E1 ∩ E2) = P(E1) . P(E2)
(As E1 and E2 are independent)
= 1/2 * 1/13 = 1/26
A. 1/6
B. 5/8
C. 3/8
D. 5/6
Explanation:
P(at least one graduate) = 1 – P(no graduates) =
1 – ⁶C₃/¹⁰C₃ = 1 – (6 * 5 * 4)/(10 * 9 * 8) = 5/6
A. 18/35
B. 12/35
C. 17/35
D. 19/35
Explanation:
If both agree stating the same fact, either both of them speak truth of both speak false.
Probability = 3/5 * 4/7 + 2/5 * 3/7
= 12/35 + 6/35 = 18/35
A. 5/21
B. 5/14
C. 9/14
D. 16/21
Explanation:
Number of ways of (selecting at least two couples among five people selected) = (⁵C₂ * ⁶C₁)
As remaining person can be any one among three couples left.
Required probability = (⁵C₂ * ⁶C₁)/¹⁰C₅
= (10 * 6)/252 = 5/21
A. 0.4869
B. 0.5904
C. 0.6234
D. 0.5834
Explanation:
P(winning prize atleast on one ticket)
= 1 – P(“Losing on all tickets”)
= 1 – (0.8)4 = (1 + (0.8)2)(1 – (0.8)2)
= (1.64)(0.36) = 0.5904
A. 1/10
B. 9/10
C. 1/5
D. 3/10
Explanation:
10 books can be rearranged in 10! ways consider the two books taken as a pair then number of favourable ways of getting these two books together is 9! 2!
Required probability = 1/5
A. 0
B. 1/7
C. 2/7
D. 5/7
Explanation:
A leap year has 52 weeks and two days
Total number of cases = 7
Number of favourable cases = 1
i.e., {Saturday, Sunday}
Required Probability = 1/7