Question

when 4 coins are tossed the probability of ) xdi). at most one tail xlv), at least two tails al most two tails xvi), at least three tails xvii). at most three tails xviii). exactly two tails xix). no head XX). no tail​

Answer 1

Answer:

When tossing four “fair” coins there are 16 equally likely outcomes, this can be found by taking the number of outcomes per event, 2; heads and tails, and raising it to the power of the number of events, 4. 2^4 = 16.

There are 6 ways to get two heads; HHTT, HTHT, HTTH, THHT, THTH, TTHH. Mathematically this is 4C2 which is equal to (4!)/[(2!)(4–2)!] =[4x3x2x1]/[(2x1)(2x1)] = 24/4 = 6

A) The probability of 2 heads is 6/16 = 3/8 = 37.5%.

To find at least two heads, you will need to add on the probability of 3 heads and also 4 heads. to your answer for exactly two heads.

The probability of 3 heads is 4/16 (or 1/4) because you have HHHT, HHTH, HTHH, THHH, or 4C3

Now the probability of 4 heads is 1/16, because the only option is HHHH.

Add 6/16 + 4/16 + 1/16 = 11/16 or 68.75%

B) The probability of at least two heads is 11/16 or 68.75%.

To find the probability of at most two heads add the probability of 1 head and no heads to your answer for exactly two heads. But because heads and tails are equally likely this is symmetrical. So P(1 Head) is the same as P(3 Heads) because 3 Heads means getting exactly one tail and getting one tail should have exactly the same probability as getting exactly one head. Likewise getting 4 heads has the same probability as 0 heads.

C) The probability of at most two heads is 11/16 or 68.75%.

Answer 2

Answer:

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Step-by-step explanation:

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