Quiz-1

Out of total of 6 occurrences there is only one possibility of getting 4 as on the dice

so the required probability will be \frac{1}{6}

In a pack of card there are a total of 13 spades card,

so the chances for being it a spade card will be

\frac{13}{52}

Atleast one tail is the non-event for all heads

Thus, P(at least 1 tail) = 1-P(all heads)

= 1 - \frac{1}{8}

= \frac{7}{8}

Odds against an event = \frac{p(E') }{ p(E)}

In this case, the event is All black, i.e; First is black and second is black and third is black

P(E) = \frac{5}{9}\times\frac{4}{8}\times\frac{ 3}{7}

=\frac{ 60}{504}

= \frac{ 5}{42}

P(E') = 1-\frac{ 5}{42}

=\frac{ 37}{42}

Odds against the event  = \frac{ 37}{5}

The event definition would be given by

first is blue and second is blue =\frac{ 3}{11} \times\frac{ 2}{10}

= \frac{3}{55}

6 \times  (\frac{4}{52}) \times (\frac{4}{51}) \times (\frac{4}{50})

= \frac{16}{5525}

P(E1) = \frac{ 3}{8}

P(E2) = \frac{7}{12}

Event definition is : E1 occurs and E2 does not occur or E1 occure and E2 occurs or E2 occurs and E1 does not occur.

(\frac{3}{8}) \times (\frac{5}{12}) + (\frac{3}{8}) \times (\frac{7}{12}) + (\frac{7}{12}) \times (\frac{5}{8})

= \frac{71}{96}

the possible outcomes are:

(1,1); (1,2) ; (2,1) ; (2,2) ; (3,1) ; (1,3)

Out of six cases in two cases there is exaclty one '2'

Thus the correct answer is \frac{2}{6}

= \frac{1}{3}

They will contradict each other if : Rajesh is true and Aman is false or vice-versa

(\frac{3}{8}) \times (\frac{1}{6}) + (\frac{5}{8}) \times  (\frac{5}{6})

=\frac{28}{48}

= \frac{7}{12}

⁴C₂ x (\frac{1}{6})^{2} \times  (\frac{5}{6})^{2}

= 6 \times  (\frac{1}{36}) \times (\frac{25}{36})

=\frac{25}{216}