When the three coins are tossed simultaneously, then the probability of getting one head will be:

Question

When the three coins are tossed simultaneously, then the probability of getting one head will be: (A)  (3/7)  (B)  (1/7)  (C)  (3/8)  (D)  (7/8)

Tardigrade Admin · Accepted Answer

Total number of cases n(S)=23=8 Favorable cases = (H,T,T),(T,H,T),(T,T,H) n(F)=3 ∴ Required probability =(3/8) Alternate Solution: Probability of getting head in one coin is p=(1/2) ⇒ q=(1/2) ∴ Probability of getting one head in three tosses = 3C1( (1/2) )1( (1/2) )2 =3( (1/2) )3 =(3/8)

Q. When the three coins are tossed simultaneously, then the probability of getting one head will be:

$\frac{3}{7}$

$\frac{1}{7}$

$\frac{3}{8}$

$\frac{7}{8}$

Q. When the three coins are tossed simultaneously, then the probability of getting one head will be:

73​

71​

83​

87​

Total number of cases n(S)=23=8 Favorable cases ={(H,T,T),(T,H,T),(T,T,H)} n(F)=3 ∴ Required probability =83​ Alternate Solution: Probability of getting head in one coin is p=21​ ⇒q=21​ ∴ Probability of getting one head in three tosses =3C1​(21​)1(21​)2 =3(21​)3 =83​

$\frac{3}{7}$

$\frac{1}{7}$

$\frac{3}{8}$

$\frac{7}{8}$