Probability Quiz 1

Solution:
P(Dhoni truth) = 0.35
P(Dhoni lie) = 0.65
P(Sachin truth) = 0.6
P(Sachin lie) = 0.4
Both Sachin’s statement and Dhoni’s statement are independent events
Case 1 - Dhoni truth and Sachin lie
= 0.35 x 0.4 = 0.14
Case 1 - Dhoni lie and Sachin truth
= 0.39

So required probability = 0.14 + 0.39
= 0.53

Solution:
Total ways of selection = 7C3 = 35
More than one vowels mean possible cases are
Case 1 - Two vowels and 1 consonant
Selection can be done in 3C2 x 4C1 = 12 ways

Case 2 - Three vowels
Selection can be done in 3C3 = 1 ways

Total possible ways = 12 + 1 = 13ways
Required probability = \frac{13}{35}

Solution:
Total number of letters = 6
No. of nonvowels = 4
P(choosing first non vowel) = \frac{4}{6}

Total number of letters remaining = 5
No. of non vowels remaining = 3
P(choosing second non vowel )= \frac{3}{5}
Required probability = \frac{4}{6}\times \frac{3}{5}
= \frac{2}{5}

Solution:
Letters common in both words are C and S
Case 1 - He chooses C
P(C from classic) = \frac{2}{7}
P(C from chess) = \frac{1}{5}
So P(C) = \frac{2}{7}\times \frac{1}{5}
= \frac{2}{35}

Case 2 - He chooses S
P(S from classic) = \frac{2}{7}
P(S from chess) = \frac{2}{5}
So P(S) = \frac{2}{7}\times \frac{2}{5}
= \frac{4}{35}

Required Probability = \frac{2}{35} + \frac{4}{35}
= \frac{6}{35}

Solution:
One person can select one option out of 3= 3C1 ways =3ways.
Hence, three persons can select one option out of 3 in 3 x 3 x 3 =27
Therefore, probability that all three choose the same option is = \frac{1}{27}

Solution:
n(S) = 25
Elements in required set E= {2,3,4,5,6,7,8,10,11,12,13,14,16,17,18,19,20,22,23,24}
n(E) = 20
Required probability = \frac{20}{25}
= \frac{4}{5}

Solution:

Number of ways of (selecting at least two couples among six people selected) = 6C2 x 8C2 = 420
(after 2 couple selecter, 2 more needed out of remaining 8)
Total ways of selection = 12C6 = 924
Required probability = \frac{420}{924}
= \frac{5}{11}

Solution:
P(at least B+) = P(B+) + P(A) + P(A+)
= 0.2 + 0.3 + 0.15 = 0.65

Solution:
P(A)=\frac{2}{7}=\frac{p}{p+q}
P(odd against A)=\frac{q}{p}
= \frac{5}{2}

Solution:
Possible outcomes of two coins tossed are {HH, HT, TH, TT}
P(TT) = \frac{1}{4}
P(6 on dice) = \frac{1}{6}
P(both to occur) = \frac{1}{4}\times \frac{1}{6}
= \frac{1}{24}