When two unbiased coins are tossed the probability of obtaining not more than 3 heads is?

Solution :  The experiment may be be taken as throwing a single coin 6 times . <br> In a single throw of a coin , we have S={H,T} <br> P(getting a head) `=(1)/(2)` and P(not getting a head) `=(1-(1)/(2)) =(1)/(2)` <br> Let X be the random varibale showing the number of heads <br> `P(X=r)=.^(n)C_(r ).p^(r ) .q^((n-r)) =.^(6)C_(r ) .((1)/(2))^(r ) .((1)/(2))^(6-r) =.^(6)C_(r ) .((1)/(2))^(6)` <br> (i) P(getting 3 heads) =P(X=)3 `=.^(6)C_(3) .((1)/(2))^(6)=(5)/(16).` <br> (ii) P(getting no head) `=P(X =0)= .^(6)C_(0).((1)/(2))^(6) =(1)/(64)` <br> (iii) P(getting at least 1 head) <br> `=1- P(X =0) =1- [.^(6)C_(0).((1)/(2))^(6)]` <br> `=(1-(1)/(64)) =(63)/(64)` <br> (iv) P(getting not more than 3 heads) <br> =P(no head or 1 head or 2 heads or 3 heads) <br> `=P(X=0) +P(X=1)+P(X=2)+P(X=3)` <br> `=.^(6)C_(0) .((1)/(2))^(6)+.^(6)C_(1).((1)/(2))^(6) +.^(6)C_(2).((1)/(6))^(6)+.^(6)C_(3).((1)/(2))^(6)` <br> `=((1)/(2))^(6) .(1+6+15+20)=((1)/(64) xx 42) =(21)/(32)`

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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