Suppose a box contains 10 coins, where three of them are type-A and seven of them are type-B. The probability of head is 0.4 if we toss a type-A coin and 0.6 for a type-B coin. One person chooses one coin at random and tosses it three times. What is the probability that this coin is type-A if the outcome is two heads out of three tosses?

0.222

Step-by-step explanation:

Given  that:

For Type A coins = 3

The probability of head in a type A coin i.e. (p) = 0.4

Then, the probability of getting a tail (q) = 1 - p = 1 - 0.4 = 0.6

For Type B coins = 7

The probability of head in a type B coin i.e. (p) = 0.6

Then, the probability of getting a tail (q) = 1 - p = 1 - 0.6 = 0.4

One person who tosses a coin three times get a probability of obtaining a head twice.

Using, the formula:

=

For Type A coin;

The probability of getting two heads in three tosses is:

= 0.288

For Type B coin;

The probability of getting two heads in three tosses is:

= 0.432

Since we have two heads out of three tosses, the probability that the coin is type A is = (P) of choosing coin A × (P) of obtaining two heads from three tosses) / total probability of getting two heads from three tosses.

However;

(P) of choosing coin A = 3/10 = 0.3

(P) of choosing coin B = 7/10 = 0.7

∴

Given that, we obtain two head from three tosses, the (P) that the coin type is A is:

= 0.222

3/4 ÷ 3/4

step-by-step explanation:

basically divide any proper fraction with itself to fit the criteria.

so if we are using 3/4, then:

3/4 ÷ 3/4 = 1

so 1 is the quotient.

and the numerators for both these fractions is 3.   and if we divide the numerators:

3 ÷ 3 = 1

we see that quotient is equal to our original quotient as well.

plz yuo

step-by-step explanation: