AP Class 10 Maths Chapter 13 Probability

In AP Class 10 Maths Chapter 13 Probability, students learn how to quantify the likelihood of an event to occur. We have listed a few important points from the chapter and provided solutions to the chapter questions here.

What is Probability?

The quantification of the likelihood of an event to occur in to a numerical measure is referred to as probability.

    • The experimental or empirical probability is Mathematically represented as follows:

[latex]P(E)=\frac{Number\, of\, trials\, in\, which\, the\, event\, happened}{Total\, number\, of\, trials}[/latex]

    • The theoretical probability (also called classical probability) of an event T, written as P(T), is defined as

[latex]P(E)=\frac{Number\, of\, outcomes\, favourable\, to\, T}{Number\, of\, all\, possible\, outcomes\, of\, the\, experiment}[/latex]

  • The probability of a sure event (or certain event) is 1 while the probability of an impossible event is 0.
  • An event with only one outcome is known as an elementary event.
  • The probability of an event E is a number P(E) such that [latex]0\leq P(E)\leq 1[/latex].
  • For any event, [latex]P(E)+P(\bar{E})=1[/latex], where [latex]\bar{E}[/latex] refers to not E.

Class 10 Maths Chapter 13 Probability Chapter Questions

    1. 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen is taken out is a good one.

Solution:

The total number of pens = 132 + 12 = 144.

The total number of good pens = 132

Hence, the probability of the taken out pen to be a good one can be calculated as

[latex]P(E)=\frac{132}{144}=\frac{11}{13}[/latex]
    1. A lot consists of 144 ball pens of which 20 are defective and the others are good. The shopkeeper draws one pen at random and gives it to Sudha. What is the probability that (i) She will buy it?

(ii) She will not buy it?

Solution:

The total number of ball pens = 144

The total number of non-defective pens = 144 – 20 = 124

(i) The probability of Sudha buying the pen is given as

[latex]P(E)=\frac{124}{144}=\frac{31}{36}[/latex]

(ii) The probability of Sudha not buying the pen is given as

[latex]P(E)=\frac{20}{144}=\frac{5}{36}[/latex]

3. One card is drawn from a well-shuffled deck of 52 cards. Calculate the probability that the card will

(i)  be an ace,

(ii) not be an ace.

Solutions: Well-shuffling ensures equally likely outcomes.
(i) There are 4 aces in a deck, so given that E is the event ‘the card is an ace’.
The number of outcomes favourable to E = 4
The number of possible outcomes = 52 (Why ?)
Therefore, P(E) =4/52= 1/13.

(ii) Now, expect F to be the event ‘card drawn is not an ace’.
The number of outcomes favourable to the event F = 52 – 4 = 48 (Why?)
The number of possible outcomes = 52
Therefore, P(F) =48 /52= 12/13

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