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In a single throw of two coins, find the probability of getting (i) both tails, (ii) at least 1 tail, (iii) at the most 1 tail.

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We know that the coin has two sides head (H) and tail (T)

So the possible outcomes are Xm. (where x is the number of outcomes when a coin is tossed and m is number of coins)

(i) When there are 2 coins∴ 22= 4 i.e. head and tail

∴ The possible outcomes are HH, HT, TH, TT.

Total possible outcomes =4

∴ Chances of getting a 2 tails = 1, i.e. TT

By using the formula,

Probability p () = number of favorable outcomes/ total number of outcomes

∴ Probability of getting a tail p (bothT) = number of two tails occurred/total number of outcomes = ¼

(ii) When there are 2 coins∴ 22= 4 i.e. head and tail

∴ The possible outcomes are HH, HT, TH, TT.

Total possible outcomes =4

∴ Chances of getting atleast 1 tail = 3, i.e. HT, TH, TT

By using the formula,

Probability p () = number of favorable outcomes/ total number of outcomes

∴ Probability of getting a tail p (atleast 1T) = number of times 1 tail occurred/total number of outcomes

=3/4

(iii) When there are 2 coins∴ 22= 4 i.e. head and tail

∴ The possible outcomes are HH, HT, TH, TT.

Total possible outcomes =4

∴ Chances of getting atmost 1 tail = 3, i.e. HT, TH, TT

By using the formula,

Probability p () = number of favorable outcomes/ total number of outcomes

∴ Probability of getting a tail p (atmost 1T) = number of times 1 tail occurred/total number of outcomes

=3/4

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