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(B) The probability of getting at least one head in $n$ tosses is given by $1 - P(	ext{no heads})$.Since the probability of getting no heads in $n$ t

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$4$

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(B) The probability of getting at least one head in $n$ tosses is given by $1 - P(	ext{no heads})$.Since the probability of getting no heads in $n$ tosses is $(1/2)^n$,the probability of getting at least one head is $1 - (1/2)^n$.We are given that this probability must be $\ge 0.9$.So,$1 - (1/2)^n \ge 0.9$.Rearranging the inequality,we get $1 - 0.9 \ge (1/2)^n$,which simplifies to $0.1 \ge (1/2)^n$.This can be written as $1/10 \ge 1/2^n$,or $2^n \ge 10$.Testing integer values for $n$:For $n = 3$,$2^3 = 8$,which is not $\ge 10$.For $n = 4$,$2^4 = 16$,which is $\ge 10$.Therefore,the minimum number of tosses required is $4$.

$Face$	$1$	$2$	$3$	$4$	$5$	$6$
$Probability$	$0.2$	$0.22$	$0.11$	$0.25$	$0.05$	$0.17$

$Face$	$1, 2, 3, 4, 5, 6$
$Probability$	$0.1, 0.24, 0.19, 0.18, 0.15, 0.14$

In order to get at least one head with a probability $\ge 0.9$,the minimum number of times a coin needs to be tossed is:

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