
A coin is tossed 5 times. What is the probability that head appears: (i) an even number of times (including 0)? (ii) an odd number of times ? (iii) at least 3 times
Answer
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Hint: Probability of an event is the ratio of Number of favorable outcomes and total number of outcomes. So note down the favorable outcomes of getting a head when a coin is tossed as per the conditions and total number of outcomes when the coin is tossed. When a coin is tossed n times then the total number of outcomes is $ {2^n} $
Complete step-by-step answer:
We are given that a coin is tossed 5 times.
So the total number of possible outcomes is $ {2^5} = 32 $ . They are
(HHHHH, HHHHT, HHHTH, HHHTT, HHTHH, HHTTH, HHTHT, HHTTT, HTHHH, HTHHT, HTHTT, HTTHH, HTTHT, HTHTH, HTTTH, HTTTT, THHHH, THHHT, THHTT, THTHH, THTTH, THTHT, THTTT, THHTH, TTHHH, TTHHT, TTHTH, TTHTT, TTTHH, TTTHT, TTTTH and TTTTT).
(i) Probability that head appears even number of items (including 0 no. of times)
Total no. of favorable outcomes that head appears even number (0, 12, 4) of times is
$ \Rightarrow {}^5{C_0} + {}^5{C_2} + {}^5{C_4} = 1 + 10 + 5 = 16 $
Total possible outcomes is 32
Probability is $ \Rightarrow \dfrac{{16}}{{32}} = \dfrac{1}{2} $
(ii) Probability that head appears odd number of times
Getting a head even number of items and getting a head odd number of times are complementary events.
If A is an event then A’ is its complementary event.
$\Rightarrow P\left( A \right) + P\left( {{A^1}} \right) = 1 $
Probability of getting head even number of times is $ \dfrac{1}{2} $
Therefore, probability of getting head odd number of times will be $ 1 - \dfrac{1}{2} = \dfrac{1}{2} $
(iii) Probability that the head appears at least 3 times.
Total no. of favorable outcomes that head appears at least 3 times is
$ \Rightarrow {}^5{C_3} + {}^5{C_4} + {}^5{C_5} = 10 + 5 + 1 = 16 $
Total possible outcomes is 32.
Probability is $ \dfrac{{16}}{{32}} = \dfrac{1}{2} $
Note: Two events are considered to be complementary, when one event occurs if and only if the other event does not occur. The probabilities of the complementary events sum up to 1. Here when the outcome is head, the outcome cannot be a tail and when the outcome is tail, the outcome cannot be head. So the events are considered as complementary.
Complete step-by-step answer:
We are given that a coin is tossed 5 times.
So the total number of possible outcomes is $ {2^5} = 32 $ . They are
(HHHHH, HHHHT, HHHTH, HHHTT, HHTHH, HHTTH, HHTHT, HHTTT, HTHHH, HTHHT, HTHTT, HTTHH, HTTHT, HTHTH, HTTTH, HTTTT, THHHH, THHHT, THHTT, THTHH, THTTH, THTHT, THTTT, THHTH, TTHHH, TTHHT, TTHTH, TTHTT, TTTHH, TTTHT, TTTTH and TTTTT).
(i) Probability that head appears even number of items (including 0 no. of times)
Total no. of favorable outcomes that head appears even number (0, 12, 4) of times is
$ \Rightarrow {}^5{C_0} + {}^5{C_2} + {}^5{C_4} = 1 + 10 + 5 = 16 $
Total possible outcomes is 32
Probability is $ \Rightarrow \dfrac{{16}}{{32}} = \dfrac{1}{2} $
(ii) Probability that head appears odd number of times
Getting a head even number of items and getting a head odd number of times are complementary events.
If A is an event then A’ is its complementary event.
$\Rightarrow P\left( A \right) + P\left( {{A^1}} \right) = 1 $
Probability of getting head even number of times is $ \dfrac{1}{2} $
Therefore, probability of getting head odd number of times will be $ 1 - \dfrac{1}{2} = \dfrac{1}{2} $
(iii) Probability that the head appears at least 3 times.
Total no. of favorable outcomes that head appears at least 3 times is
$ \Rightarrow {}^5{C_3} + {}^5{C_4} + {}^5{C_5} = 10 + 5 + 1 = 16 $
Total possible outcomes is 32.
Probability is $ \dfrac{{16}}{{32}} = \dfrac{1}{2} $
Note: Two events are considered to be complementary, when one event occurs if and only if the other event does not occur. The probabilities of the complementary events sum up to 1. Here when the outcome is head, the outcome cannot be a tail and when the outcome is tail, the outcome cannot be head. So the events are considered as complementary.
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