Exercise 13.2

In a match between A and B: \[\left( \mathbf{i} \right)\] the probability of winning of A is \[\mathbf{0}.\mathbf{83}\]. What is the probability of winning of B? \[\left( \mathbf{ii} \right)\] the probability of losing the match is \[\mathbf{0}.\mathbf{49}\] for B. What is the probability of winning of A?

Solution: \[\left( i \right)\]We know that, The probability of winning of A \[+\]Probability of losing of A \[=\text{ }1\] And, Probability of losing of A \[=\] Probability of winning of B...

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From a well shuffled deck of \[\mathbf{52}\] cards, one card is drawn. Find the probability that the card drawn is: \[\left( \mathbf{v} \right)\] a card with number less than \[\mathbf{8}\] \[\left( \mathbf{vi} \right)\] a card with number between \[\mathbf{2}\] and \[\mathbf{9}\]

Solution: \[\left( v \right)\] Numbers less than \[8\text{ }=\text{ }\left\{ \text{ }2,\text{ }3,\text{ }4,\text{ }5,\text{ }6,\text{ }7 \right\}\]\[\] Event of drawing a card with number less than...

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From a well shuffled deck of \[\mathbf{52}\] cards, one card is drawn. Find the probability that the card drawn is: \[\left( \mathbf{iii} \right)\]a queen of black card \[\left( \mathbf{iv} \right)\]a card with number \[\mathbf{5}\text{ }\mathbf{or}\text{ }\mathbf{6}\]

  Solution: \[\left( iii \right)\] Event of drawing a queen of black colour \[=\text{ }\left\{ Q\left( spade \right),\text{ }Q\left( club \right) \right\}\text{ }=\text{ }E\] So,\[~n\left( E...

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A bag contains \[\mathbf{3}\] red balls, \[\mathbf{4}\] blue balls and \[\mathbf{1}\] yellow ball, all the balls being identical in shape and size. If a ball is taken out of the bag without looking into it; find the probability that the ball is: \[\left( \mathbf{iii} \right)\]not yellow \[\left( \mathbf{iv} \right)\] neither yellow nor red

Solution: \[\left( iii \right)\] Probability of not drawing a yellow ball \[=\text{ }1\text{ }\] Probability of drawing a yellow ball Thus, probability of not drawing a yellow ball \[=\text{...

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A bag contains \[\mathbf{3}\] red balls, \[\mathbf{4}\] blue balls and \[\mathbf{1}\] yellow ball, all the balls being identical in shape and size. If a ball is taken out of the bag without looking into it; find the probability that the ball is: \[\left( \mathbf{i} \right)\] yellow \[\left( \mathbf{ii} \right)\] red

Solution: The total number of balls in the bag \[=\text{ }3\text{ }+\text{ }4\text{ }+\text{ }1\text{ }=\text{ }8\] balls So, the number of possible outcomes \[=\text{ }8\text{ }=\text{ }n\left( S...

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From identical cards, numbered one card is drawn at random. Find the probability that the number on the card drawn is a multiple of: \[\left( \mathbf{iii} \right)\text{ }\mathbf{3}\text{ }\mathbf{and}\text{ }\mathbf{5}\] \[\left( \mathbf{iv} \right)\text{ }\mathbf{3}\text{ }\mathbf{or}\text{ }\mathbf{5}\]

Solution: \[\left( iii \right)\] From numbers \[1\text{ }to\text{ }25\], there is only one number which is multiple of \[3\text{ }and\text{ }5\text{ }i.e.~\left\{ 15 \right\}\] So, favorable number...

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From \[\mathbf{25}\]identical cards, \[\text{ }\mathbf{1},\text{ }\mathbf{2},\text{ }\mathbf{3},\text{ }\mathbf{4},\text{ }\mathbf{5},\text{ }\ldots \ldots ,~\mathbf{24},\text{ }\mathbf{25}:\]numbered one card is drawn at random. Find the probability that the number on the card drawn is a multiple of\[\left( \mathbf{i} \right)\text{ }\mathbf{3}\] \[\left( \mathbf{ii} \right)\text{ }\mathbf{5}\]

Solution: We know that, there are \[25\] cards from which one card is drawn. So, the total number of elementary events \[=\text{ }n\left( S \right)\text{ }=\text{ }25\] \[\left( i \right)\]From...

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$Hundred identical cards are numbered from \[\mathbf{1}\text{ }\mathbf{to}\text{ }\mathbf{100}\] The cards The cards are well shuffled and then a card is drawn. Find the probability that the number on card drawn is: \[\left( \mathbf{v} \right)\] less than \[\mathbf{48}\]

Solution: \[\left( v \right)\]From numbers \[1\text{ }to\text{ }100\], there are \[47\] numbers which are less than \[48\text{ }i.e.~\{1,\text{ }2,\text{ }\ldots \ldots \ldots ..,\]\[46,\text{...

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Hundred identical cards are numbered from The cards The cards are well shuffled and then a card is drawn. Find the probability that the number on card drawn is: \[\left( \mathbf{iii} \right)\] between \[\mathbf{40}\] and \[\mathbf{60}\] \[\left( \mathbf{iv} \right)\] greater than \[\mathbf{85}\]

Solution: \[\left( iii \right)\] From numbers \[1\text{ }to\text{ }100\], there are \[19\] numbers which are between \[40\text{ }and\text{ }60\text{ }i.e.~\{41,\text{ }42\], \[43,\text{ }44,\text{...

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Hundred identical cards are numbered from \[\mathbf{1}\text{ }\mathbf{to}\text{ }\mathbf{100}\]. The cards The cards are well shuffled and then a card is drawn. Find the probability that the number on card drawn is: \[\left( \mathbf{i} \right)\] a multiple of \[\mathbf{5}\] \[\left( \mathbf{ii} \right)\] a multiple of \[\mathbf{6}\]

Solution: We kwon that, there are \[100\] cards from which one card is drawn. Total number of elementary events \[=\text{ }n\left( S \right)\text{ }=\text{ }100\] \[\left( i \right)~\] From numbers...

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multiple Nine cards (identical in all respects) are numbered . A card is selected from them at random. Find the probability that the card selected will be: \[\left( \mathbf{iii} \right)\] an even number and a multiple of \[\mathbf{3}\] \[\left( \mathbf{iv} \right)\] an even number or a of \[\mathbf{3}\]

Solution: \[\left( iii \right)\] From numbers \[2\text{ }to\text{ }10\], there is one number which is an even number as well as multiple of \[3\text{ }i.e.\text{ }6\] So, favorable number of events...

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Nine cards (identical in all respects) are numbered \[\mathbf{2}\text{ }\mathbf{to}\text{ }\mathbf{10}\]. A card is selected from them at random. Find the probability that the card selected will be: \[\left( \mathbf{i} \right)\]an even number \[\left( \mathbf{ii} \right)\] a multiple of \[\mathbf{3}\]

Solution: We know that, there are totally \[9\] cards from which one card is drawn. Total number of elementary events \[=\text{ }n\left( S \right)\text{ }=\text{ }9\] \[\left( i \right)\] From...

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In calculating the mean of grouped data, grouped in classes of equal width, we may use the formula where a is the assumed mean. a must be one of the mid-points of the classes. Is the last statement correct? Justify your answer.

No, the assertion isn't right. It isn't required that expected mean ought to be the mid – mark of the class span. a can be considered as any worth which is not difficult to work on it.

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Apoorv throws two dice once and computes the product of the numbers appearing on the dice. Peehu throws one die and squares the number that appears on it. Who has the better chance of getting the number 36? Why?

Apoorv toss two dice on the double. Thus, the all out number of results = 36 Number of results for getting item 36 = 1(6×6) ∴ Probability for Apoorv = 1/36 Peehu tosses one kick the bucket, Thus,...

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