Ap ssc probability exercise 1
Exercise-1
Example-1. Find the probability of getting a head when a coin is tossed once. Also find the probability of getting a tail.
Solution :In the experiment of tossing a coin once, the number of possible outcomes is two.They are Head (H) and Tail (T).
Let E be the event 'getting a head'. The number of outcomes favourable to E is 1.
Therefore,
P(E) = P (head) =`\frac{Number of outcomes favourabLe To E}{Number of all possibLe outcomes }` =`\frac{1}{2}`
P(E) = P (tail) =`\frac{Number Of Outcomes FavourabLe To F}{Number of all PossibLe Outcomes }` =`\frac{1}{2}`
Example-2. A bag contains a red ball, a blue ball and an yellow ball, all the balls being of the same size. Manasa takes out a ball from the bag without looking into it. What is the probability that she takes a (i) yellow ball? (ii) red ball? (iii) blue ball?
Solution :Manasa takes out a ball from the bag without looking into it. So, it is equally likely that she takes out any one of them.
Now, the number of possible outcomes = 3.
The number of outcomes favourable to the event YellowBall = 1.
So, P(YelloBall) =`\frac{Number of outcomes favourabLe To YelloBall}{Number of all possibLe outcomes }` =`\frac{1}{3}`
Similarly, P(RedBall) =`\frac{1}{3}`
and P(BlueBall) =`\frac{1}{3}`
Note:
In above two examples all the events are elementary events.So we get sum of the events=1
- P(head)+P(tail)=1
- P(YellowBall)+P(RedBall)+P(BlueBall)=1
Example-3. Suppose we throw a die once. (i) What is the probability of getting a number greater than 4? (ii) What is the probability of getting a number less than or equal to 4?
Solution :(i) In rolling an unbaised dice
Sample space S = {1, 2, 3, 4, 5, 6}
No. of outcomes n(S) = 6
Favourable outcomes for E = {5, 6}
number greater than 4
No. of favourable outcomes n(E) = 2
So, P(E) =`\frac{Number of outcomes favourabLe To E}{Number of all possibLe outcomes }`
=`\frac{2}{6}`=`\frac{1}{3}`
(ii) Let F be the event 'getting a number less than or equal to 4'.
Sample space S = {1, 2, 3, 4, 5, 6}
No. of outcomes n(S) = 6
Favourable outcomes for F = {1, 2, 3, 4}
number less than or equal to 4
No. of favourable outcomes n(F) = 4
So, P(F) =`\frac{Number of outcomes favourabLe To F}{Number of all possibLe outcomes }`
=`\frac{4}{6}`=`\frac{2}{3}`
Example-4. 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.
Solution :Well-shuffling ensures equally likely outcomes.
(i) There are 4 aces in a deck.
The number of outcomes favourable to Ace = 4
The number of possible outcomes = 52
`\therefore` P(Ace) =`\frac{Number of outcomes favourabLe To Ace}{Number of all possibLe outcomes }`
=`\frac{4}{52}`=`\frac{1}{13}`
(ii) The number of outcomes favourable to the event not an Ace = 52 - 4 = 48
The number of possible outcomes = 52
`\therefore` P(not an Ace) =`\frac{Number of outcomes favourabLe To Not an Ace}{Number of all possibLe outcomes }`
=`\frac{48}{52}`=`\frac{12}{13}`
Short Method: P(not an Ace)= 1-`\frac{1}{13}`=`\frac{12}{13}` [`\because`P(`\bar{E}` ) = 1 – P(E)]
Example-5. Sangeeta and Reshma, play a tennis match. It is known that the probability of Sangeeta winning the match is 0.62. What is the probability of Reshma winning the match?
Solution :Let S and R denote the events that Sangeeta wins the match and Reshma wins the match, respectively.
The probability of Sangeeta's winning chances = P(S) = 0.62 (given)
The probability of Reshma's winning chances = P(R) = 1 - P(S)
= 1 -0.62 = 0.38 [`\because`R and S are complementary]
Example-6. Sarada and Hamida are friends. What is the probability that both will have (i) different birthdays? (ii) the same birthday? (ignoring a leap year).
Solution :Out of the two friends, one girl, say, Sarada's birthday can be any day of the year. Now, Hamida's birthday can also be any day of 365 days in the year. We assume that these 365 outcomes are equally likely.
(i) If Hamida's birthday is different from Sarada's,
the number of favourable outcomes for Hamida birthday is 365 - 1 = 364
So, P (Hamida's birthday is different from Sarada's birthday) =`\frac{364}{365}`
(ii)P(Sarada and Hamida have the same birthday) = 1 - P (both have different birthdays)
=1-`\frac{364}{365}`=`\frac{1}{365}`
Example-7. There are 40 students in Class X of a school of whom 25 are girls and 15 are boys. The class teacher has to select one student as a class representative. She writes the name of each student on a separate cards, the cards being identical. Then she puts cards in a box and stirs them thoroughly. She then draws one card from the box. What is the probability that the name written on the card is the name of (i) a girl? (ii) a boy?
Solution :There are 40 students, and only one name card has to be chosen. The number of all possible outcomes is 40
(i) The number of outcomes favourable for a card with the name of a girl = 25
`\therefore` P (card with name of a girl) = P(Girl) =`\frac{25}{40}`=`\frac{5}{8}`
(ii) The number of outcomes favourable for a card with the name of a boy = 15
`\therefore` P(card with name of a boy) = P(Boy) =`\frac{15}{40}`=`\frac{3}{8}`
OrP(Boy) = 1 - P(not Boy) = 1 - P(Girl) = 1 -`\frac{5}{8}`=`\frac{3}{8}`
(i). 1 (ii) 0, Impossible event (iii) 1 , Sure event (iv) 1 (v) 0 ,1
(i)No.Generally car starts.But its depend on condition of car so its not equally likely outcome.
(ii)No. Generally he doesnot miss the shoot rare cases he may miss so its not equally likely out .
(iii)Yes.This trial have only two cases that answer may be right or it may be wrong.
(iv) Yes.The chances for a boy or a girl are same.
P(E) is given as 0.05
P(`\bar{E}`) = 1 – P(E) = 1 – 0.05 = 0.95
`\therefore`P(`\bar{E}`) = 0.95
(i) There is no lemon flavoured candy in the bag so The probability that she takes out an orange flavoured candy is 0.
(ii)The bag has only lemon flovoured candies.so this is certain event So the probability that she takes out a lemon flavoured candy is 1.
Number cards in the deck is 52.
After removing all hearts from deck, number of cards remained in deck =52 -13 =39
`\therefore`The number of possible outcomes =39.
(i) After removing heart cards number of favourable outcomes to Ace=3
Probability of getting an ace from the remaining pack = P(Ace) =`\frac{Number of outcomes favourabLe To Ace}{Number of all possibLe outcomes } = `\frac{3}{39}`=`\frac{1}{13}`
(ii)Favourable outcomes to Diamond card=13
Probability of getting a diamond from the remaining pack =P(diamond)=`\frac{13}{39}`=`\frac{1}{3}`
(iii) Heart cards are already removed from the deck.So remaining 39 cards are not heart cards.
So this is sure event.P(Not Heart card)=`\frac{39}{39}`=1
(iv) After removing all heart cards there is no ace heart card in the deck
So this is an impossible event.So P( Ace of hearts) = `\frac{0}{39}`=0
P(Not having same birthday) = 0.992
Probability of same birthday P(Same birthday)=1- P(Not having same birthday) =1-0.992=0.008
(i) Total no of possible outcomes = 6 {1, 2, 3, 4, 5, 6}
E ⟶ Event of getting a prime number.
Number of favorable outcomes = 3 {2, 3, 5}
P(E) =`\frac{Number of favourabLe outcomes}{Number of all possibLe outcomes }
P(E) =`\frac{3}{6}`=`\frac{1}{2}`
(ii) E ⟶ Event of getting a number lying between 2 and 6.
Number of favorable outcomes = 3 {3, 4, 5}
Total number of possible outcomes = 6
P(E) =`\frac{3}{6}`=`\frac{1}{2}`
(iii) E ⟶ Event of getting an Odd number.
Number of favorable outcomes = 3 {1,3,5}
Total number of possible outcomes = 6 {1, 2, 3, 4, 5, 6}
P(E) =`\frac{3}{6}`=`\frac{1}{2}`
Total number of outcomes = 52 {52 cards}
E⟶ event of getting a red king
No of favourable outcomes = 2{king of hearts & king of diamonds}
We know that, P(E) =`\frac{Number of favourabLe outcomes}{Number of all possibLe outcomes }
P(E) =`\frac{2}{52}`=`\frac{1}{26}`
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