Let a biased coin be tossed 5 times. If the probability of getting 4 heads is equal to the probability of getting 5 heads, then the probability of getting atmost two heads is:

Question

Let a biased coin be tossed 5 times. If the probability of getting 4 heads is equal to the probability of getting 5 heads, then the probability of getting atmost two heads is: (A) (275/65) (B) (36/54) (C) (181/55) (D) (46/64)

Tardigrade Admin · Accepted Answer

P ( H )= x , P ( T )=1- x P (4 H , 1 T )= P (5 H ) 5 C 1( x )4(1- x )1= 5 C 5 x 5 5(1- x )= x 6 x =5=0 x =(5/6) P ( atmost 2 H ) = P ( OH , 5 T )+ P (1 H , 4 T )+ P (2 H , 3 T ) = 5 C 0((1/6))5+ 5 C 1 (5/6) ⋅((1/6))4+ 5 C 2((5/6))3((1/6))3 =(1/65)(1+25+250)=(276/65) = (46/64)

Q. Let a biased coin be tossed $5$ times. If the probability of getting $4$ heads is equal to the probability of getting $5$ heads, then the probability of getting atmost two heads is:

$\frac{275}{6 ^{5}}$

$\frac{36}{5 ^{4}}$

$\frac{181}{5 ^{5}}$

$\frac{46}{6 ^{4}}$

Q. Let a biased coin be tossed 5 times. If the probability of getting 4 heads is equal to the probability of getting 5 heads, then the probability of getting atmost two heads is:

65275​

5436​

55181​

6446​

P(H)=x,P(T)=1−x P(4H,1T)=P(5H) 5C1​(x)4(1−x)1=5C5​x5 5(1−x)=x 6x=5=0x=65​ P( atmost 2H) =P(OH,5T)+P(1H,4T)+P(2H,3T) =5C0​(61​)5+5C1​65​⋅(61​)4+5C2​(65​)3(61​)3 =651​(1+25+250)=65276​ =6446​

Q. Let a biased coin be tossed $5$ times. If the probability of getting $4$ heads is equal to the probability of getting $5$ heads, then the probability of getting atmost two heads is:

$\frac{275}{6 ^{5}}$

$\frac{36}{5 ^{4}}$

$\frac{181}{5 ^{5}}$

$\frac{46}{6 ^{4}}$