RD Chapter 33 Probability Ex 33.3 Solutions
Question - 11 : - A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is:
(i) a black king
(ii) either a black card or a king
(iii) black and a king
(iv) a jack, queen or a king
(v) neither an ace nor a king
(vi) spade or an ace
(vii) neither an ace nor a king
(viii) a diamond card
(ix) not a diamond card
(x) a black card
(xi) not an ace
(xii) not a black card
Answer - 11 : -
Given: Pack of 52cards.
By using the formula,
P (E) = favourableoutcomes / total possible outcomes
We know that, a cardis drawn from a pack of 52 cards, so number of elementary events in the samplespace is
n (S) = 52C1 =52
(i) Let E be theevent of drawing a black king
n (E) =2C1 =2(there are two black kings one of spade and other of club)
P (E) = n (E) / n (S)
= 2 / 52
= 1/26
(ii) Let E be theevent of drawing a black card or a king
n (E) = 26C1+4C1–2C1=28
[Weare subtracting 2 from total because there are two black king which are alreadycounted and to avoid the error of considering it twice.]
P (E) = n (E) / n (S)
= 28 / 52
= 7/13
(iii) Let E be theevent of drawing a black card and a king
n (E) =2C1 =2 (there are two black kings one of spade and other of club)
P (E) = n (E) / n (S)
= 2 / 52
= 1/26
(iv) Let E be theevent of drawing a jack, queen or king
n (E) = 4C1+4C1+4C1 =12
P (E) = n (E) / n (S)
= 12 / 52
= 3/13
(v) Let E be theevent of drawing neither a heart nor a king
Now let us consider E′as the event that either a heart or king appears
n (E′) = 6C1+4C1-1=16(there is a heart king so it is deducted)
P (E′) = n (E′) / n(S)
= 16 / 52
= 4/13
So, P (E) = 1 – P (E′)
= 1 – 4/13
= 9/13
(vi) Let E be theevent of drawing a spade or king
n (E)=13C1+4C1-1=16
P (E) = n (E) / n (S)
= 16 / 52
= 4/13
(vii) Let E be theevent of drawing neither an ace nor a king
Now let us consider E′as the event that either an ace or king appears
n(E′) = 4C1+4C1 =8
P (E′) = n (E′) / n(S)
= 8 / 52
= 2/13
So, P (E) = 1 – P (E′)
= 1 – 2/13
= 11/13
(viii) Let E be theevent of drawing a diamond card
n (E)=13C1=13
P (E) = n (E) / n (S)
= 13 / 52
= ¼
(ix) Let E be theevent of drawing not a diamond card
Now let us consider E′as the event that diamond card appears
n (E′) =13C1=13
P (E′) = n (E′) / n(S)
= 13 / 52
= 1/4
So, P (E) = 1 – P (E′)
= 1 – 1/4
= ¾
(x) Let E be theevent of drawing a black card
n (E) =26C1 =26 (spades and clubs)
P (E) = n (E) / n (S)
= 26 / 52
= ½
(xi) Let E be theevent of drawing not an ace
Now let us consider E′as the event that ace card appears
n (E′) = 4C1 =4
P (E′) = n (E′) / n(S)
= 4 / 52
= 1/13
So, P (E) = 1 – P (E′)
= 1 – 1/13
=12/13
(xii) Let E be theevent of not drawing a black card
n (E) = 26C1 =26 (red cards of hearts and diamonds)
P (E) = n (E) / n (S)
= 26 / 52
= ½
Question - 12 : - In shutting a pack of 52 playing cards, four are accidently dropped; find the chance that the missing cards should be one from each suit
Answer - 12 : -
Given: A pack of 52cards from which 4 are dropped.
By using the formula,
P (E) = favourableoutcomes / total possible outcomes
We now have to findthe probability that the missing cards should be one from each suit
We know that, fromwell shuffled pack of cards, 4 cards missed out total possible outcomes are
n (S) = 52C4 =270725
Let E be the eventthat four missing cards are from each suite
n (E) = 13C1×13C1×13C1×13C1 =134
P (E) = n (E) / n (S)
= 134 /270725
= 2197/20825
Question - 13 : - From a deck of 52 cards, four cards are drawn simultaneously, find the chance that they will be the four honors of the same suit
Answer - 13 : -
Given: A deck of 52 cards.
By using the formula,
P (E) = favourableoutcomes / total possible outcomes
We now have to findthe probability that all the face cards of same suits are drawn.
Total possibleoutcomes are
n (S) = 52C4
Let E be the eventthat all the cards drawn are face cards of same suit.
n (E)=4×4C4=4
P (E) = n (E) / n (S)
= 4 / 270725
Question - 14 : - Tickets numbered from 1 to 20 are mixed up together and then a ticket is drawn at random. What is the probability that the ticket has a number which is a multiple of 3 or 7?
Answer - 14 : -
Given: Numberedtickets from 1 to 20.
By using the formula,
P (E) = favourableoutcomes / total possible outcomes
To find theprobability of the ticket drawn having a number which is a multiple of 3 or 7.
We know that, oneticket is drawn from a lot of mixed number.
Total possibleoutcomes are
n (S) = 20C1 =20
Let E be the event ofgetting ticket which has number that is multiple of 3 or 7
E ={3,6,9,12,15,18,7,14}
n (E) = 8
P (E) = n (E) / n (S)
= 8 / 20
= 2/5
Question - 15 : - A bag contains 6 red, 4 white and 8 blue balls. If three balls are drawn at random, find the probability that one is red, one is white and one is blue
Answer - 15 : -
Given: A bagcontaining 6 red, 4 white and 8 blue balls.
By using the formula,
P (E) = favourableoutcomes / total possible outcomes
Three balls are drawnso, we have to find the probability that one is red, one is white and one isblue.
Total number ofoutcomes for drawing 3 balls is 18C3
n (S) = 18C3 =816
Let E be the eventthat one red, one white and one blue ball is drawn.
n (E) = 6C14C18C1 =192
P (E) = n (E) / n (S)
= 192 / 816
= 4/17
Question - 16 : - A bag contains 7 white, 5 black and 4 red balls. If two balls are drawn at random, find the probability that:
(i) both the balls are white
(ii) one ball is black and the other red
(iii) both the balls are of the same colour
Answer - 16 : -
Given: A bagcontaining 7 white, 5 black and 4 red balls.
By using the formula,
P (E) = favourableoutcomes / total possible outcomes
Two balls are drawn atrandom, therefore
Total possibleoutcomes are 16C2
n (S) = 120
(i) Let E be theevent of getting both white balls
E = {(W) (W)}
n (E) = 7C2 =21
P (E) = n (E) / n (S)
= 21 / 120
= 7/40
(ii) Let E be theevent of getting one black and one red ball
E = {(B) (R)}
n (E) = 5C14C1 =20
P (E) = n (E) / n (S)
= 20 / 120
= 1/6
(iii) Let E be theevent of getting both balls of same colour
E = {(B) (B)} or {(W)(W)} or {(R) (R)}
n (E) = 7C2+5C2+4C2 =37
P (E) = n (E) / n (S)
= 37 / 120
Question - 17 : - A bag contains 6 red, 4 white and 8 blue balls. If three balls are drawn at random, find the probability that:
(i) one is red and two are white
(ii) two are blue and one is red
(iii) one is red
Answer - 17 : -
Given: A bagcontaining 6 red, 4 white and 8 blue balls.
By using the formula,
P (E) = favourableoutcomes / total possible outcomes
Two balls are drawn atrandom.
Total possibleoutcomes are 18C3
n (S) = 816
(i) Let E be theevent of getting one red and two white balls
E = {(W) (W) (R)}
n (E) = 6C14C2 =36
P (E) = n (E) / n (S)
= 36 / 816
= 3/68
(ii) Let E be theevent of getting two blue and one red
E = {(B) (B) (R)}
n (E) = 8C26C1 =168
P (E) = n (E) / n (S)
= 168 / 816
= 7/34
(iii) Let E be theevent that one of the balls must be red
E = {(R) (B) (B)} or{(R) (W) (W)} or {(R) (B) (W)}
n (E) = 6C14C18C1+6C14C2+6C18C2 =396
P (E) = n (E) / n (S)
= 396 / 816
= 33/68
Question - 18 : - Five cards are drawn from a pack of 52 cards. What is the chance that these 5 will contain:
(i) just one ace
(ii) at least one ace?
Answer - 18 : -
Given: Five cards aredrawn from a pack of 52 cards.
By using the formula,
P (E) = favourableoutcomes / total possible outcomes
Five cards are drawnat random,
Total possibleoutcomes are 52C5
n (S) = 2598960
(i) Let E be theevent that exactly only one ace is present
n (E) = 4C148C4 =778320
P (E) = n (E) / n (S)
= 778320 / 2598960
= 3243/10829
(ii) Let E be theevent that at least one ace is present
E = {1 or 2 or 3 or 4ace(s)}
n (E) = 4C148C4+4C248C3+4C348C2+4C448C1 =886656
P (E) = n (E) / n (S)
= 886656 / 2598960
= 18472/54145
Question - 19 : - The face cards are removed from a full pack. Out of the remaining 40 cards, 4 are drawn at random. What is the probability that they belong to different suits?
Answer - 19 : -
Given: The face cardsare removed from a full pack of 52.
By using the formula,
P (E) = favourableoutcomes / total possible outcomes
Four cards are drawnfrom the remaining 40 cards, so we have to find the probability that all ofthem belong to different suit.
Total possibleoutcomes of drawing four cards are 40C4
n (S) = 40C4 =91390
Let E be the eventthat 4 cards belong to different suit.
n (E) = 10C110C110C110C1 =10000
P (E) = n (E) / n (S)
= 10000 / 91390
= 1000/9139
Question - 20 : - There are four men and six women on the city councils. If one council member is selected for a committee at random, how likely is that it is a women?
Answer - 20 : -
Given: There are fourmen and six women on the city councils.
By using the formula,
P (E) = favourableoutcomes / total possible outcomes
From the city councilone person is selected as a council member so, we have to find the probabilitythat it is a woman.
Total possibleoutcomes of selecting a person is 10C1
n (S)= 10C1 =10
Let E be the eventthat it is a woman
n (E) = 6C1 =6
P (E) = n (E) / n (S)
= 6 / 10
= 3/5