Answer
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Hint: If 3 coins are tossed then sample space will be S= {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT}. To find the probability of an event use the formula, P(event) = (No. of favourable outcomes) / (Total no. of possible outcomes).
Complete step-by-step answer:
Given in the question that 3 coins are tossed simultaneously,
${\text{a}}{\text{.)}}$The possible outcomes are-
$S = \{ HHH, HHT, HTT, HTH, THT, TTT, THH, TTH\} $
${\text{b}}{\text{.)}}$ As we know the possible outcomes are-
$S = \{ HHH, HHT, HTT, HTH, THT, TTT, THH, TTH\} $,
So, the total no. of possible outcomes will be 8.
${\text{c}}{\text{.)}}$ Now, favourable outcomes of getting at least one head are {HHH, HHT, HTT, THT, THH, TTH, HTH}.
So, the total no. of favourable outcomes is 7.
And the total no. of possible outcomes is 8.
Using the formula, P(event) = (No. of favourable outcomes) / (Total no. of possible outcomes), we get-
P (getting at least one head) $ = \dfrac{7}{8}$
${\text{d}}{\text{.)}}$Now, favourable outcomes of getting at most two heads (getting two or less than two heads) are {HHT, HTT, TTT, THH, HTH, THT, TTH}.
So, the total no. of favourable outcomes is 7.
And the total no. of possible outcomes is 8.
Using the formula, P(event) = (No. of favourable outcomes) / (Total no. of possible outcomes), we get-
P (getting at most two head) $ = \dfrac{7}{8}$
${\text{e}}{\text{.)}}$ Now, favourable outcomes of getting no tails are {HHH}.
So, the total no. of favourable outcomes is 1.
And the total no. of possible outcomes is 8.
Using the formula, P(event) = (No. of favourable outcomes) / (Total no. of possible outcomes), we get-
P (getting no tails) $ = \dfrac{1}{8}$
Note: Whenever such types of questions appear, then always write the sample space of all outcomes, then for each problem type write the favourable outcomes for that given event. And then solve and find the answers to each part.
Complete step-by-step answer:
Given in the question that 3 coins are tossed simultaneously,
${\text{a}}{\text{.)}}$The possible outcomes are-
$S = \{ HHH, HHT, HTT, HTH, THT, TTT, THH, TTH\} $
${\text{b}}{\text{.)}}$ As we know the possible outcomes are-
$S = \{ HHH, HHT, HTT, HTH, THT, TTT, THH, TTH\} $,
So, the total no. of possible outcomes will be 8.
${\text{c}}{\text{.)}}$ Now, favourable outcomes of getting at least one head are {HHH, HHT, HTT, THT, THH, TTH, HTH}.
So, the total no. of favourable outcomes is 7.
And the total no. of possible outcomes is 8.
Using the formula, P(event) = (No. of favourable outcomes) / (Total no. of possible outcomes), we get-
P (getting at least one head) $ = \dfrac{7}{8}$
${\text{d}}{\text{.)}}$Now, favourable outcomes of getting at most two heads (getting two or less than two heads) are {HHT, HTT, TTT, THH, HTH, THT, TTH}.
So, the total no. of favourable outcomes is 7.
And the total no. of possible outcomes is 8.
Using the formula, P(event) = (No. of favourable outcomes) / (Total no. of possible outcomes), we get-
P (getting at most two head) $ = \dfrac{7}{8}$
${\text{e}}{\text{.)}}$ Now, favourable outcomes of getting no tails are {HHH}.
So, the total no. of favourable outcomes is 1.
And the total no. of possible outcomes is 8.
Using the formula, P(event) = (No. of favourable outcomes) / (Total no. of possible outcomes), we get-
P (getting no tails) $ = \dfrac{1}{8}$
Note: Whenever such types of questions appear, then always write the sample space of all outcomes, then for each problem type write the favourable outcomes for that given event. And then solve and find the answers to each part.
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