How To Solve Probability Questions Quickly

Type 2- How to Solve Probability Questions Quickly of boys and girls

Question 1. In a class there are 60% of girls of which 25% poor. What is the probability that a poor girl is selected is leader?

Options

(a) 20%

(b) 15%

(c) 10%

(d) 25%

Solutions:   Assume total students in the class = 100

Then Girls = 60% (100) = 60

Poor girls = 25% (60) = 15

So probability that a poor girls is selected leader = \frac{Poor   girls}{ Total  students} = \frac{15}{100} = 15%

Correct Options (b)

Questions 2. What is the probability that the total of two dice will be greater than 9, given that the first die is a 5?

Options

(a) \frac{1}{3}

(b) \frac{1}{6}

(c) \frac{1}{9}

(d) None of these

Solution :    Let A= first die is 5

Let B = total of two dice is greater than 9

P(A) = Possible outcomes for A and B: (5, 5), (5, 6)

P(A and B) = \frac{2}{36} = \frac{1}{18}

P(B|A) = \frac{P(A  and   B)}{P(A)} = \frac{1}{18} ÷ \frac{1}{6} = \frac{1}{3}.

Correct Options (A)

Questions 3. If six cards are selected at random (without replacement) from a standard deck of 52 cards, what is the probability there will be no pairs? (two cards of the same denomination)

Solution:    Let E_i be the event that the first i cards have no pair among them. Then we want to compute P(E₆).

Which is actually the same as P(E₁ ∩ E₂ ∩ · · · ∩ E₆), since E₆ ⊂ E₅ ⊂ · · · ⊂ E₁, implying that E₁ ∩ E₂ ∩ · · · ∩ E₆ = E₆.

We get P(E₁ ∩ E₂ ∩ · · · ∩ E₆) = P(E₁)P(E₂|E₁)· · ·  = \frac{52}{52} \frac{48}{51} \frac{44}{50} \frac{40}{49} \frac{36}{48} \frac{32}{47}

Alternatively, one can solve the problem directly using counting techniques.

Define the sample space to be (equally likely)ordered sequences of 6 cards; then, |S| = 52 · 51 · 50 · · · 47, and the event E6 has 52 · 48 · 44 · · · 32 elements

Questions 4.  A group of 5 friends-Archie, Betty, Jerry, Moose, and Veronica-arrived at the movie theater to see a movie. Because they arrived late, their only seating option consists of 3 middle seats in the front row, an aisle seat in the front row, and an adjoining seat in the third row. If Archie, Jerry, or Moose must sit in the aisle seat while Betty and Veronica refuse to sit next to each other, how many possible seating arrangements are there?

Options

(a) 32 

(b) 36 

(c) 48 

(d) 72 

(e) 120

Solution:     Good = Total – Bad. 

Total = arrangements with Archie, Jerry or Moose in the aisle seat.

Number of options for the aisle seat = 3. (Archie, Jughead, or Moose) 

Number of ways to arrange the 4 other people = 4  x 3 x 2 x 1. 

To combine these options, we multiply:  3 x 4 x 3 x 2 = 72. 

Bad = arrangements with Archie, Jerry or Moose in the aisle seat BUT with Betty next to Veronica.

Number of options for the aisle seat = 3. (Archie, Jughead, Moose). 

Number of options for the third row seat = 2. (Anyone but Betty and Veronica, since in a bad arrangement they sit next to each other.) 

Number of options for the middle of the 3 remaining seats = 2. (Must be Betty or Veronica so that they sit next to each other). 

Number of ways to arrange the 2 remaining people = 2 x 1. 

To combine these options, we multiply:  3 x 2 x 2 x 2 = 24. 

Good arrangements = 72 – 24 = 48. 

Correct Option C.