Solution: option (1) 1/9 If two dice are thrown simultaneously, the total number of sample space is 36 Favourable outcomes = (1, 4), (4, 1), (2, 3) and (3, 2) Therefore, the required probability = 4/ 36 = 1/9.
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`1/6`
Explanation:
When two dice are thrown simultaneously, all possible outcomes are:
(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)
Number of all possible outcomes = 36
Let E be the event of getting a doublet.
Then the favourable outcomes are:
(1,1), (2,2) , (3,3) (4,4), (5,5), (6,6)
Number of favourable outcomes = 6
∴ P(getting a doublet) = P ( E) = `6/36 = 1/6`
Page 2
`3/4`
Explanation:
When two coins are tossed simultaneously, the possible outcomes are HH, HT, TH and TT.
Total number of possible outcomes = 4
Let E be the event of getting at most one head.
Then, the favourable outcomes are HT, TH and TT.
Number of favourable outcomes = 3
∴ P(getting at most 1head) = `("Number of favourable outcomes")/"Number of all possible outcomes" = 3/4`
Page 3
`3/8`
Explanation:
When 3 coins are tossed simultaneously, the possible outcomes are HHH, HHT, HTH, THH, THT, HTT, TTH and TTT.
Total number of possible outcomes = 8
Let E be the event of getting exactly two heads.
Then, the favourable outcomes are HHT, THH, and HTH.
Number of favourable outcomes = 3
∴ Probability of getting exactly 2 heads = P(E) =`3/8`
Select the correct option from the given alternatives :
Two dice are thrown simultaneously. Then the probability of getting two numbers whose product is even is
`3/4`
Explanation;
Two dice are thrown.
∴ n(S) = 36.
Getting two numbers whose product is even, i.e., one of the two numbers must be even.
Let event A: Getting even number on first dice.
event B: Getting even number on second dice.
∴ n(A) = 18, n(B) = 18, n(A ∩ B) = 9
∴ Required probability = P(A ∩ B)
= `("n"("A") + "n"("B") - "n"("A" ∩ "B"))/("n"("S"))`
= `(18 + 18 - 9)/36`
= `27/36`
= `3/4`
Concept: Concept of Probability
Is there an error in this question or solution?
Correct Answer: D
Solution :
| Two dice are thrown simultaneously |
| Total number of outcomes = 36 |
| Doublet = {(1, 1), (2, 2), (3, 3), (4, 4), (5,5), (6, 6)} |
| Total of atleast 10 = {(4, 6), (5, 5), (6, 4), (5, 6), (6, 5), (6, 6)} |
| Total is perfect square = {(1, 3), (2, 2), (3, 1), (3, 6), (4, 5), (5,4), (6, 3)} |
| [a] P (doublet) =6/36 =1/6, |
| P (not doublet) \[=1-\frac{1}{6}=\frac{5}{6}\] |
| [b] P (getting total of atleast 10) |
| =6/36=1/6 |
| [c] P (perfect square) =7/36, |
| P (not a perfect square) \[=1-\frac{7}{36}=\frac{29}{36}\] |
Page 2
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Correct Answer: D
Solution :
Total number of outcomes \[=6\times 6\times 6=216\] Number of favourable outcomes \[=6\times 5\times 4=120\] \[\therefore \]Required probability \[=\frac{120}{216}=\frac{5}{9}\]
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Page 3
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Correct Answer: A
Solution :
Total face cards (king, queen and jack) \[=3\times 4=12\] \[P\left( Face\,card \right)=\frac{Total\,Face\,Cards}{Total\,cards}\] \[=\frac{12}{52}=\frac{3}{13}\]
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Page 4
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Correct Answer: A
Solution :
In a deck of 52 cards, there are 12 face cards i.e., 6 red and 6 black cards. \[\therefore \]So, probability of getting a red face card \[=\frac{6}{52}=\frac{3}{26}\]
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Page 5
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Correct Answer: D
Solution :
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Page 6
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Correct Answer: D
Solution :
Total number of cards in one deck of cards, \[n\left( S \right)=52\] Let \[{{E}_{1}}\]= Event of getting a king of red colour \[\therefore \,\,\,\,n\left( {{E}_{1}} \right)=2\] (\[\because \]In a deck of cards, 26 cards are red and 26 cards are black. There are four kings in a deck in which two are red and two are black) Probability of getting a king of red colour \[=\frac{n\left( {{E}_{1}} \right)}{n\left( S \right)}=\frac{2}{52}=\frac{1}{26}\]
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Page 7
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Correct Answer: B
Solution :
Total number of cards = 52 Total number of diamond cards =13 [a] P (diamond cards) = 13 / 52 = 1 /4 [b] P (an ace of heart) =1/52 [c] P (not heart) \[=1-\frac{1}{4}=\frac{3}{4}\] [d] P (king or queen) \[=\frac{4}{52}+\frac{4}{52}=\frac{8}{52}=\frac{2}{13}\]
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Page 8
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Correct Answer: C
Solution :
Total number of outcomes = 52 Number of favourable outcomes = 2 \[P\left( E \right)=\frac{2}{52}=\frac{1}{26}\]
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Page 9
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Correct Answer: A
Solution :
Number of red balls = 5 Let the number of blue balls in the bag = x \[\therefore \]Total number of balls = x + 5 \[\therefore \]Probability of red ball \[=\frac{5}{5+x}\] and probability (blue balls) \[=\frac{x}{5+x}\] According to question, \[\frac{x}{5+x}=2\times \frac{5}{5+x}\] x = 10
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Page 10
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Correct Answer: C
Solution :
Let there be x blue balls in the bag. \[\therefore \]Total number of balls in the bag \[=\left( \text{ }8\text{ }+\text{ }x \right)\]and Now, \[{{P}_{1}}\]= Probability of drawing a blue ball \[=\frac{x}{8+x}\]and \[{{P}_{2}}\]= Probability of drawing a red ball \[=\frac{8}{8+x}\] It is given that, \[{{P}_{1}}=3{{P}_{2}}\] \[\Rightarrow \,\,\,\,\,\frac{x}{8+x}=3\times \frac{8}{\left( 8+x \right)}\] \[\Rightarrow \,\,\,\frac{x}{8+x}=\frac{24}{8+x}\] Hence, there are 24 blue balls in the bag.
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Page 11
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Correct Answer: D
Solution :
Total number of balls in a bag =3+5 \[\therefore \] Total number of possible outcomes is 15. The number of favourable outcomes of drawing non-black ball is 10. \[\therefore \]Required probability \[=\frac{Number\text{ }of\text{ }favourable\text{ }outcomes}{Total\text{ }number\text{ }of\text{ }possible\text{ }outcomes}=\frac{10}{15}=\frac{2}{3}\]
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Page 12
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Correct Answer: B
Solution :
Let the number of defective bulbs = x Probability of getting a defective bulb \[=\frac{x}{500}\] According to question, \[\frac{x}{500}=0.290\] \[\Rightarrow \,\,x=0.290\times 500=145\]
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Page 13
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Correct Answer: C
Solution :
Prime number between 1 to 100 =2,3,5,7, 11, 13, 17, 19,23,29 31,37,41, 43, 47, 53, 59, 61 67, 71, 73, 79, 83, 89, 97 = 25 Total outcomes =100 \[P\left( E \right)=\frac{25}{100}=\frac{1}{4}\]
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Page 14
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Correct Answer: D
Solution :
Clearly, number x can take any one of the five given values. So, total number of possible outcomes = 5. We observe that \[{{x}^{2}}<2\] when x takes any one of the following three values -1, 0 and 1. So, favourable number of elementary events = 3 Hence, \[P\left( {{x}^{2}}<2 \right)=\frac{3}{5}\]
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Page 15
Correct Answer: C
Solution :
Number x can be selected in three ways and corresponding to each such way there are three ways of selecting number y. Therefore, two numbers can be selected in 9 ways as listed below (1,1), (1,4), (1,9), (2,1), (2,4), (2,9), (3,1), (3,4), (3,9) So, total numbers of possible outcomes = 9 The product xy will be less than 9, if x and y are chosen in one of the following ways (1, 1), (1,4), (2,1), (2,4), (3.1) \[\therefore \]Favourable number of elementary events = 5 Hence, required probability \[=\frac{5}{9}\]Page 16
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Correct Answer: B
Solution :
As given that a number other than six has appeared. So, the man repeating it to be six means he is speaking false. Effectively the question is asking the probability P (He will lie) \[=1-\frac{3}{4}=\frac{1}{4}\] Hence, probability that he reports it is a six is \[\frac{1}{4}\]
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Page 17
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Correct Answer: B
Solution :
Total number of envelopes in the box =1000 Number of envelopes containing cash prize =10+100+200=310 Number of envelopes containing no cash prize = 1000 - 310 = 690 \[\therefore \]Required probability \[=\frac{690}{1000}=0.69\]
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Page 18
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Correct Answer: A
Solution :
Out of 20 tickets numbered from 1 to 20, one can be chosen in 20 ways. So, total number of possible outcomes associated with the given random experiment is 20. Out of 20 tickets numbered 1 to 20, tickets bearing numbers which are multiple of 3 or 7 bear numbers 3, 6, 7, 9, 12, 14, 15 and 18. \[\therefore \]Favourable number of elementary events =8 Hence, required probability \[=\frac{8}{20}=\frac{2}{5}\]
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Page 19
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Correct Answer: B
Solution :
There are 14 = (5 + 9) fish out of which one can be chosen in 14 ways. \[\therefore \]Total numbers of possible outcomes =14 There are 5 male fish out of which one male fish can be chosen in 5 ways. \[\therefore \]Favourable number of elementary events = 5 Hence, required probability \[=\frac{5}{14}\]
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Page 20
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Correct Answer: B
Solution :
Let boys be B and girl be G Outcomes can be BBB, GGG, BBG, BGB, GBB, GGB, GBG, BGG Then, probability of 3 girls \[=\frac{1}{8}\] Probability of 0 girls \[=\frac{1}{8}\] Probability of 2 girls \[=\frac{3}{8}\] Probability of 1 girls \[=\frac{3}{8}\]
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Page 21
Correct Answer: B
Solution :
[A] Probability of getting number 5 in throwing a dice \[=\frac{1}{6}\] [B] Probability of getting three heads in a single throw of coin = 0 [C] Probability of getting the sum of the number as 7 [(3, 4), (4, 3), (1, 6), (6, 1), (2, 5), (5, 2)] when two dice are thrown \[=\frac{6}{36}\] [D] Probability of occurence of two sure independent events is \[\frac{2}{7}\].Page 22
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Correct Answer: A
Solution :
[A] The probability of sure event is 1. [B] The probability of impossible event is 0. [C] Number of face cards in the pack of cards is \[\left( 3\times 4 \right)=12\] [D] Probability of occuring 53 Sundays in a leap year is \[\frac{2}{7}\].
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