Probability — Analytical & Application Based Questions

A bag contains 13 red cards, 13 black cards and 13 green cards. Each set of cards are numbered 1 to 13. From these cards, a card is drawn at random. What is the probability that the card drawn is a:

(a) green card?

(b) a card with an even number?

(c) a red or black card with a number which is a multiple of three?

Answer

Total no. of cards = 13 + 13 + 13 = 39 cards.

(a) P(that card drawn is a green card) = $\dfrac{\text{No. of green cards}}{\text{Total no. of cards}} = \dfrac{13}{39} = \dfrac{1}{3}$.

Hence, probability that card drawn is a green card = $\dfrac{1}{3}$.

(b) Even number cards are : 2, 4, 6, 8, 10, 12.

So, there are 6 even cards of each set.

∴ 18 cards.

P(that card drawn is a card with even number)

= $\dfrac{\text{No. of even number cards}}{\text{Total no. of cards}} = \dfrac{18}{39} = \dfrac{6}{13}$.

Hence, probability that card drawn is a card with even number = $\dfrac{6}{13}$.

(c) Multiples of three : 3, 6, 9, 12.

So, there are 4 cards of each red and black colour.

∴ 8 cards.

P(that card drawn is a red or black card with multiple of three)

= $\dfrac{\text{No. of red or black card with multiple of 3}}{\text{Total no. of cards}} = \dfrac{8}{39}$.

Hence, probability that card drawn is a red or black card with multiple of three = $\dfrac{8}{39}$.

The probability of selecting a blue marble and a red marble from a bag containing red, blue and green marbles is $\dfrac{1}{3}$ and $\dfrac{1}{5}$ respectively. If the bag contains 14 green marbles, then find :

(a) number of red marbles.

(b) total number of marbles in the bag.

Answer

As the bag contains red, blue and green marbles.

∴ Probability of selecting a red marble + Probability of selecting a blue marble + Probability of selecting a green marble = 1

⇒ $\dfrac{1}{3} + \dfrac{1}{5}$ + Probability of selecting a green marble = 1

⇒ $\dfrac{5 + 3}{15}$ + Probability of selecting a green marble = 1

⇒ Probability of selecting a green marble = $1 - \dfrac{8}{15}$

⇒ Probability of selecting a green marble = $\dfrac{15 - 8}{15}$

⇒ Probability of selecting a green marble = $\dfrac{7}{15}$.

$\therefore \dfrac{\text{No. of green marbles}}{\text{Total no. of marbles}} = \dfrac{7}{15} \\[1em] \Rightarrow \dfrac{14}{\text{Total no. of marbles}} = \dfrac{7}{15} \\[1em] \Rightarrow \text{Total no. of marbles} = \dfrac{15 \times 14}{7} = 30.$

(a) Given,

⇒ Probability of selecting a red marble = $\dfrac{1}{5}$.

$\therefore \dfrac{\text{No. of red marbles}}{\text{Total no. of marbles}} = \dfrac{1}{5} \\[1em] \Rightarrow \dfrac{\text{No. of red marbles}}{30} = \dfrac{1}{5} \\[1em] \Rightarrow \text{No. of red marbles} = \dfrac{30}{5} = 6.$

Hence, no. of red marbles = 6.

(b) Hence, total no. of marbles in the bag = 30.

The marks scored by 100 students are given below:

A student in the class is selected at random. Find the probability that the student has scored:

(a) less than 20

(b) below 60 but 30 or more

(c) more than or equal to 70

(d) above 89.

Answer

(a) By formula,

Probability that student has scored less than 20

= $\dfrac{\text{No. of students who scored less than 20}}{\text{Total no. of students}} = \dfrac{9}{100}$.

Hence, probability that the student has scored less than 20 = $\dfrac{9}{100}$.

(b) From table,

No. of students who scored less than 60 = 50

No. of students who scored less than 30 = 18

∴ No. of students who score below 60 but 30 or more = 50 - 18 = 32.

Probability that student has scored below 60 but 30 or more = $\dfrac{\text{No. of students scoring between 60 and 30}}{\text{Total no. of students}} = \dfrac{32}{100} = \dfrac{8}{25}$.

Hence, probability that the student has scored below 60 but 30 or more = $\dfrac{8}{25}$.

(c) From table,

No. of students who scored less than 70 = 65

Total no. of students = 100

∴ No. of students who score more than or equal to 70 = 100 - 65 = 35.

Probability that student has scored more than or equal to 70 = $\dfrac{\text{No. of students scoring 70 or more}}{\text{Total no. of students}} = \dfrac{35}{100} = \dfrac{7}{20}$.

Hence, probability that the student has scored more than or equal to 70 = $\dfrac{7}{20}$.

(d) No. of students those who have scored more than 89 = No. of students who has scored between 90-100 = 10.

Probability that student has scored more than 89 = $\dfrac{\text{No. of students scoring \textgreater 89}}{\text{Total no. of students}} = \dfrac{10}{100} = \dfrac{1}{10}$.

Hence, probability that the student has scored more than 89 = $\dfrac{1}{10}$.