Question

Two Biased coins are thrown together. On one of the coins head appears
twice as frequently as tail on the other for every 2 times
that head appears, tail appear
s 3 times. The probability that
the coins show both heads or both tail is ??​

Answer 1

Answer:

See below

Step-by-step explanation:

When we toss two coins simultaneously then the possible of outcomes are: (two heads) or (one head and one tail) or (two tails) i.e., in short (H, H) or (H, T) or (T, T) respectively; where H is denoted for head and T is denoted for tail.

Answer 2

ADDITIONAL INFORMATION:-

When we toss two coins simultaneously then the possible of outcomes are: (two heads) or (one head and one tail) or (two tails) i.e., in short (H, H) or (H, T) or (T, T) respectively; where H is denoted for head and T is denoted for tail.

When we toss two coins simultaneously then the possible of outcomes are: (two heads) or (one head and one tail) or (two tails) i.e., in short (H, H) or (H, T) or (T, T) respectively; where H is denoted for head and T is denoted for tail.Therefore, total numbers of outcome are 22 = 4

When we toss two coins simultaneously then the possible of outcomes are: (two heads) or (one head and one tail) or (two tails) i.e., in short (H, H) or (H, T) or (T, T) respectively; where H is denoted for head and T is denoted for tail.Therefore, total numbers of outcome are 22 = 4The above explanation will help us to solve the problems on finding the probability of tossing two coins.

SOLUTION:-

When two different coins are tossed randomly, the sample space is given by

When two different coins are tossed randomly, the sample space is given byS = {HH, HT, TH, TT}

When two different coins are tossed randomly, the sample space is given byS = {HH, HT, TH, TT}Therefore, n(S) = 4.

• getting two heads:

Let E1 = event of getting 2 heads. Then, E1 = {HH} and, therefore,

n(E1) = 1.

Therefore, P(getting 2 heads) = P(E1) = n(E1)/n(S) = 1/4.

• getting two tails:
• getting two tails:Let E2 = event of getting 2 tails. Then,
• getting two tails:Let E2 = event of getting 2 tails. Then, E2 = {TT} and, therefore, n(E2) = 1.
• getting two tails:Let E2 = event of getting 2 tails. Then, E2 = {TT} and, therefore, n(E2) = 1. Therefore, P(getting 2 tails) = P(E2) = n(E2)/n(S) = 1/4.