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Probability

Probability

MODULES

Basics of Probability
Mutually Exclusive Events
Non-Mutually Exclusive Events
Independent Events
Dependent Events
Conditional Probability
Odds For & Against
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Probability : Level 1
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ALL MODULES

CAT 2025 Lesson : Probability - Mutually Exclusive Events

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3. Coins, Dice and Cards

In a lot of entrance tests, we are expected to understand tossing a coin, rolling a die (or dice) or picking a card from a standard deck of cards. Sometimes the questions do not explain these events in detail. Therefore, we have explained these below.

111) Tossing a coin: A standard coin has two sides to it. One is called Heads and the other is called Tails.
In an unbiased coin, these 2\bm{2}2 outcomes are possible and both of them are equally likely.

222) Rolling a die: A standard die is a cube with 666 sides. Each side is uniquely numbered – 1,2,3,4,5\bm{1, 2, 3, 4, 5}1,2,3,4,5 and 6\bm{6}6. These 666 outcomes are equally likely when a die is rolled. (Note that dice is the plural for die.)

333) Standard deck of cards: There are a total of 525252 cards in a standard deck.
Ranks: There are 131313 ranks. 999 of them are numbered from 222 to 101010, followed by Jack (J),
Queen (Q), King (K) and Ace (A).
The 131313 cards stated in order are −-− A, 2,3,4,5,6,7,8,9,102, 3, 4, 5, 6, 7, 8, 9, 102,3,4,5,6,7,8,9,10, J, Q, K
Suits: There are 444 suits – Clubs, Diamonds, Hearts and Spades. Each of these suits contain the
131313 differently-ranked cards mentioned above. Therefore, a deck contains 4×13=524 \times 13 = 524×13=52 cards
Colour: Each of the cards are either Red or Black. The colour is based on the suit of the card. Clubs and Spades are black-coloured cards while Diamonds and Hearts are red-coloured cards.
Therefore, there are 262626 black cards and 262626 red cards in a standard deck.

4. Exclusivity of Events

4.1 Mutually Exclusive Events

When 222 events are such that at most one of them can occur at a time and not both, they are called mutually exclusive events. For instance, when we toss a coin, getting a head or a tail are mutually exclusive events.

In other words, two or more events are mutually exclusive if there is no common outcome between them, i.e., P(A ∩\cap∩ B) =0= \bm{0}=0

Probability of occurrence of either of the two mutually exclusive events (A or B) is P(A or B) === P(A ∪\cup∪ B) === P(A) + P(B)

Example 4

There are 333 black balls, 444 red balls and 666 white balls. What is the probability of drawing a black or red ball?

(111) 313\dfrac{3}{13}133​               (222) 613\dfrac{6}{13}136​                 (3) 713\dfrac{7}{13}137​               (4) 913\dfrac{9}{13}139​             

Solution

Total number of outcomes =n(S)=3+4+6=13= n(\text{S}) = 3 + 4 + 6 = 13=n(S)=3+4+6=13
No. of favourable outcomes =n(E)=3+4=7= n(\text{E}) = 3 + 4 = 7=n(E)=3+4=7

P(E) =n(E)n(S)=713 = \dfrac{n(\text{E})}{n\text(S)} = \dfrac{7}{13}=n(S)n(E)​=137​

Answer: (333) 713\dfrac{7}{13}137​

Example 5

Four coins are tossed simultaneously. What is the probability of
(I) getting a head in at least two of the coins?
(II) getting a head in at least one of the coins?

Solution

Case I: Each coin has 222 outcomes – head or a tail. Imagine the 444 coins being tossed
simultaneously by 444 people standing in a row. Each of these coins can have 222 outcomes.
Total outcomes when 444 coins are tossed =2×2×2×2=16= 2 \times 2 \times 2 \times 2 = 16=2×2×2×2=16

Favourable outcomes are 222 heads & 222 tails, 333 heads & 111 tails and 444 heads & no tails.
Let H denote heads and T denote tails. Then, HHTT signifies the results of the coins in that order – coins of the first two people returning heads and the last two returning tails. Number of different such outcomes with 222 heads and 222 tails is the number of ways letters in the word HHTT can be permuted.

Number of permutations for HHTT =4!2!×2!=6= \dfrac{4 !}{2 ! \times 2 !} = 6=2!×2!4!​=6 ; HHHT =4!3!=4= \dfrac{4 !}{3 !} = 4=3!4!​=4 ; HHHH =4!4!=1= \dfrac{4 !}{4 !} = 1=4!4!​=1

Probability =6+4+116=1116= \dfrac{6 + 4 + 1}{16} = \dfrac{\bm{11}}{\bm{16}}=166+4+1​=1611​

Case II: We shall first find the probability of getting a tail in all coins and then subtract this from 111.

Number of permutations for TTTT =4!4!=1= \dfrac{4 !}{4 !} = 1=4!4!​=1

Probability of getting TTTT =116= \dfrac{1}{16}=161​

Probability of getting at least 111 head =1−116=1516= 1 - \dfrac{1}{16} = \dfrac{\bm{15}}{\bm{16}}=1−161​=1615​

Answer: (I) 1116\dfrac{11}{16}1611​ ;     (II) 1516\dfrac{15}{16}1615​

Example 6

When two six-sided dice are rolled, what is the probability of the sum of the numbers on the two dice being greater than 999?

Solution

As there are two dice and each of which have 666 outcomes,
Total outcomes when two dice are rolled =6×6=36= 6 \times 6 = 36=6×6=36

Favourable outcomes are when the sum of the number are 10,1110, 1110,11 or 121212
101010 ⇒ {(6,4),(5,5),(4,6)(6, 4), (5, 5), (4, 6)(6,4),(5,5),(4,6)}
111111 ⇒ {(6,5),(5,6)(6, 5), (5, 6)(6,5),(5,6)}
121212 ⇒ {(6,6)(6, 6)(6,6)}

No. of favourable outcomes =6= 6=6

Probability =636=16= \dfrac{6}{36} = \dfrac{1}{6}=366​=61​

Answer: 16\dfrac{1}{6}61​

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