Ex 14.1, 10 - A piggy bank contains hundred 50 p coins, fifty - Ex 15.1

part 2 - Ex 14.1, 10 - Ex 15.1 - Serial order wise - Chapter 14 Class 10 Probability

part 3 - Ex 14.1, 10 - Ex 15.1 - Serial order wise - Chapter 14 Class 10 Probability part 4 - Ex 14.1, 10 - Ex 15.1 - Serial order wise - Chapter 14 Class 10 Probability

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Ex 14.1, 10 A piggy bank contains hundred 50 p coins, fifty Rs 1 coins, twenty Rs 2 coins and ten Rs 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin will be a 50 p coin? Total number of coins in a piggy bank = 100 + 50 + 20 + 10 = 180 Number of 50 p coins = 100 Probability of getting a 50 p coin = (๐‘๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘๐‘œ๐‘–๐‘›๐‘  ๐‘œ๐‘“ 50๐‘)/(๐‘‡๐‘œ๐‘ก๐‘Ž๐‘™ ๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘๐‘œ๐‘–๐‘›๐‘ ) = 100/180 = ๐Ÿ“/๐Ÿ— Ex 14.1, 10 A piggy bank contains hundred 50 p coins, fifty Rs 1 coins, twenty Rs 2 coins and ten Rs 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin (ii) will not be a โ‚น 5 coin? Total number of coins in a piggy bank = 180 Number of โ‚น 5 coins = 10 P(not getting a โ‚น 5 coin) = 1 โ€“ P(getting a โ‚น 5 coin) Now, P(getting a โ‚น 5 coin) = (๐‘๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘๐‘œ๐‘–๐‘›๐‘  ๐‘œ๐‘“ โ‚น 5 )/(๐‘‡๐‘œ๐‘ก๐‘Ž๐‘™ ๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘๐‘œ๐‘–๐‘›๐‘ ) = 10/180 = ๐Ÿ/๐Ÿ๐Ÿ– Thus, P(not getting a โ‚น 5 coin) = 1 โ€“ P(getting a โ‚น 5 coin) = 1 โ€“ 1/18 = ๐Ÿ๐Ÿ•/๐Ÿ๐Ÿ–

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