Question

A coin is tossed three times. Find the probability of the following events.
(1) A: getting at least two heads
(2) B: getting exactly two heads.
(3) C: getting at most one head.

Answer 1

Hint: A coin is tossed three times, so the sample space:$\\{ HHH,HTH,THH,TTH,HHT,HTT,THT,TTT\\}$, so, total number of possible outcomes = 8. Use the formula, $P(event) = \dfrac{{No.ofFavourableOutcomes}}{{TotalNo.ofPossibleOutcomes}}$, where number of favourable outcomes means the outcomes which may take place for that event and total number of possible outcomes means the total possibilities that may occur when a coin is tossed three times. According to given cases find the number of favourable outcomes and probability of that particular case.

Complete step-by-step answer:
(1) A: getting at least two heads
$P(A) = P(GettingTwoHeads) + P(GettingThreeHeads)$
Now for getting two heads, the number of favourable outcomes will be 3 which are $\\{HTH, THH, HHT\\}$.
And the total no. of possible outcomes is 8 which are $\\{ HHH,HTH,THH,TTH,HHT,HTT,THT,TTT\\}$.
Now for getting three heads, no. of favourable outcomes will be 1 which is $\\{HHH\\}$.
Therefore, probability is-
$P(A) = \dfrac{3}{8} + \dfrac{1}{8} = \dfrac{4}{8} = \dfrac{1}{2}$ (using formula $P(event) = \dfrac{{No.of Favourable Outcomes}}{{TotalNo.ofPossibleOutcomes}}$)
$\therefore P(A) = \dfrac{1}{2}$

(2) B: getting exactly two heads
Now for getting two heads, no. of favourable outcomes will be 3 which are $\\{HTH, THH, HHT\\}$.
And the total no. of possible outcomes is 8 which are $\\{ HHH,HTH,THH,TTH,HHT,HTT,THT,TTT\\}$.
Therefore, the probability is-
$P(B) = P(GettingExactlyTwoHeads) \\\ P(B) = \dfrac{3}{8} \\\$ (using formula $P(event) = \dfrac{{No.ofFavourableOutcomes}}{{TotalNo.ofPossibleOutcomes}}$)

(3) C: getting at most one head.
Getting atmost one head means number of heads can be 0 or 1 i.e getting no head and one head

Now for getting no heads, no. of favourable outcomes will be 1 which is $\\{TTT\\}$.
And the total no. of possible outcomes is 8 which are $\\{ HHH,HTH,THH,TTH,HHT,HTT,THT,TTT\\}$.
Now for getting one head, no. of favourable outcomes will be 4 which is $\\{HTT, THT, TTT\\}$.
Therefore, probability is-
(using formula $P(event) = \dfrac{{No.ofFavourableOutcomes}}{{TotalNo.ofPossibleOutcomes}}$)
$P(C) = P(GettingNoHeads) + P(GettingOneHeads) \\\ P(C) = \dfrac{1}{8} + \dfrac{3}{8} = \dfrac{4}{8} = \dfrac{1}{2} \\\ \therefore P(C) = \dfrac{1}{2} \\\$

Note- Whenever such a type of question appears note down all the outcomes of the event and the possible outcomes in the particular given case, as given in question the coin is tossed 3 times. And then find the probability of all the cases using the standard formula.