Two unbiased coins are tossed, what is the probability of getting at most one head

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One coin are tossed getting probability = head or tailwhere tail probability = 1/2when two coin tossed getting probability = head or tail , tail or headwhere tail probability = 2/4=1/2option A right hai

7

Two unbiased coins are tossed simultaneously. Find the probability of getting at most one tail.

The outcomes When two unbiased coins are tossed simultaneously:

HH, HT, TH, TT

No. of favorable outcomes = 3

No. of total outcomes = 4

As we know,

Probability =

⇒ Probability (Getting at most one tail)

=

Answer

Verified

Hint- In order to solve such type of question we must use formula Probability \[\left( {\text{P}} \right){\text{ = }}\dfrac{{{\text{Favourable number of cases}}}}{{{\text{Total number of cases}}}}\] , along with proper understanding of favourable cases and total cases.

Complete step-by-step answer:
If two unbiased coins are tossed simultaneously, then the total number of possible outcomes may be either.
1.both head HH
2.one head and one tail (HT, TH)
3.Both tail (TT)

So here total number of possible cases = 4

(i) two heads
a favourable outcome for two heads is HH.
No. of favourable outcomes = 1
total number of possible cases = 4

As we know that Probability \[\left( {\text{P}} \right){\text{ = }}\dfrac{{{\text{Favourable number of cases}}}}{{{\text{Total number of cases}}}}\]
Probability (two heads) =$\dfrac{1}{4}$

(ii) favourable outcomes for one tail are TH, HT.
No. of favourable outcomes = 2
total number of possible cases = 4

As we know that Probability \[\left( {\text{P}} \right){\text{ = }}\dfrac{{{\text{Favourable number of cases}}}}{{{\text{Total number of cases}}}}\]

Probability(one tail) = $\dfrac{2}{4}{\text{ = }}\dfrac{1}{2}$

(iii) favourable outcomes for at least one tail are TH, HT, TT.
No. of favourable outcomes = 3
total number of possible cases = 4

As we know that Probability \[\left( {\text{P}} \right){\text{ = }}\dfrac{{{\text{Favourable number of cases}}}}{{{\text{Total number of cases}}}}\]

Probability (at least one tail) = $\dfrac{3}{4}$

(iv) Favourable outcomes for at least one head are TH, HT, HH.
No. of favourable outcomes = 3
total number of possible cases = 4

As we know that Probability \[\left( {\text{P}} \right){\text{ = }}\dfrac{{{\text{Favourable number of cases}}}}{{{\text{Total number of cases}}}}\]

Probability (at least one head) = $\dfrac{3}{4}$

(v) favourable outcomes for no tail means not a single head is HH.
No. of favourable outcomes = 1
total number of possible cases = 4

As we know that Probability \[\left( {\text{P}} \right){\text{ = }}\dfrac{{{\text{Favourable number of cases}}}}{{{\text{Total number of cases}}}}\]

Probability(no tail) = $\dfrac{1}{4}$

NOTE- In Such Types of Question first find out the total numbers of possible outcomes then find out the number of favourable cases, then divide them using the formula which stated above, we will get the required answer.

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